Question 13 Marks
Justify the following statements.
- Reactions with $\Delta\text{G}<0$ always have an equilibrium constant greater than 1.
- Many thermodynamically feasible reactions do not occur under ordinary conditions.
- At low temperatures, enthalpy change dominates the $\Delta\text{G}$ expression and at high temperatures, it is the entropy which dominates the value of $\Delta\text{G}.$
Answer
- $\Delta\text{G}=-2.303\text{RT}\log\text{K}.$ Thus, when $\Delta\text{G}^\circ<0,$ K > 1).
- Under ordinary conditions, the average energy of the reactants may be less than threshold energy. They require some activation energy to initiate the reaction.
- $\Delta\text{G}=\Delta\text{H}-\text{T}\Delta\text{S}.$ At low temperature, $\text{T}\Delta\text{S}$ is small. Hence, $\Delta\text{H}$ dominates. At high temperature, $\text{T}\Delta\text{S}$ is large, i.e. $\Delta\text{S}$ dominates the value of $\Delta\text{G}.$
View full question & answer→Question 23 Marks
Although heat is a path function but heat absorbed by the system under certain specific conditions is independent of path. What are those conditions? Explain.
AnswerHeat is independent of path under two conditions :
- At constant volume
By first law of thermodynamics,
$\text{q}=\Delta\text{U}+(-\text{w})$
and, $(-\text{w})=\text{P}\Delta\text{V}$
Therefore, $\text{q}=\Delta\text{U}+\text{P}\Delta\text{V}$
By first law of thermodynamics $\Delta\text{V}=0$
or, $\text{qV}=\Delta\text{U}+0$
or, $\text{qV}=\Delta\text{U}=$ change in internal energy
- At constant pressure
$\text{q}_\text{p}=\Delta\text{U}+\text{P}\Delta\text{V}$
But, $\Delta\text{U}+\text{P}\Delta\text{V}=\Delta\text{H}$
Therefore, $\text{q}_\text{p}=\Delta\text{H}=$ change in enthalpy
So, at constant volume and at constant pressure heat change is a state function because it is equal to change in internal energy and change in enthalpy respectively which are state functions. View full question & answer→Question 33 Marks
- Why entropy of steam is more than that of water at its boiling point?
- Out of diamond and graphite which has higher entropy?
- Standard heat of formation of hydrazine $\left[\mathrm{N}_2 \mathrm{H}_4(\mathrm{l})\right.$, hydrogen peroxide $\left[\mathrm{H}_2 \mathrm{O}_2(\mathrm{l})\right]$ and water $\left[\mathrm{H}_2 \mathrm{O}(\mathrm{l})\right]$ are $-50.4,-193.2$ and $-242.7 \mathrm{~kJ} / \mathrm{mole}$ respectively. Calculate the standard heat of formation for the following reaction:
$\text{N}_2\text{H}_4\text{(l)}+2\text{H}_2\text{O}_2(\text{l})\overrightarrow{\ \ \ \ \ \ \ }\ \text{N}_2(\text{g})+4\text{H}_2\text{O(l)}$Answer
- Steam has more disorder or randomness as compared to liquid water, therefore, has more entropy.
- Graphite has more entropy than diamond because force of attraction between layers is less in graphite than diamond.
- $\text{N}_2\text{H}_4\text{(l)}+2\text{H}_2\text{O}_2(\text{l})\overrightarrow{\ \ \ \ \ \ \ }\ \text{N}_2(\text{g})+4\text{H}_2\text{O(l)}$
$\Delta_\text{r}\text{H}=\Delta_\text{f}\text{H}(\text{N}_2)+4\Delta_\text{f}\text{H}(\text{H}_2\text{O})\\-\Delta_\text{f}\text{H}(\text{N}_2\text{H}_4)-2\Delta_\text{f}\text{H}(\text{H}_2\text{O})$
$=0+4\times[-242.7-(-50.4)-2(-193.2)]$
$=(-970.8+50.4+386.4)\text{kJ mol}^{-1}$
$\Delta_\text{r}\text{H}=-534\text{kJ mol}^{-1}$ View full question & answer→Question 43 Marks
Differentiate between the following (with examples)
- Open and closed system.
- Adiabatic and isothermal process.
- State function and path function.
Answer
- Open system: Open system is a system which can exchange both matter as well as energy e.g. a cup of tea.
Closed system: Closed system is a system which can exchange energy but not matter e.g. tea placed in closed kettle.
- Adiabatic process: Adiabatic process is a process in which no exchange of heat takes place with the surrounding e.g. carrying out reaction in a isolated system. The conductor should have non-conducting walls.
Isothermal process: Isothermal process is a process in which no change in temperature takes place e.g. thermostat maintains constant temperature by exchanging heat with the surroundings.
- State function: It depends upon initial and final state of the system and not on path e.g. $\Delta\text{U}$ (internal energy change)
Path function: It depends upon path e.g. work. View full question & answer→Question 53 Marks
Calculate $\Delta_{\mathrm{f}} \mathrm{H}^{\circ}$ of HCl if bond energy of $\mathrm{H}-\mathrm{H}$ bond is $437 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{Cl}-\mathrm{Cl}$ bond is $244 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\mathrm{H}-\mathrm{Cl}$ is $433 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
Answer$\frac{1}{2}\text{H}_2(\text{g})+\frac{1}{2}\text{Cl}_2(\text{g})\overrightarrow{\ \ \ \ \ \ } \ \text{HCl(g)}$$\Delta_\text{f}\text{H}^\circ=\sum$ Bond enthalpy of reactants $-\sum$ Bond enthalpy of products.
$=\frac{1}{2}\text{B}_{\text{H}-\text{H}}+\frac{1}{2}\text{B}_{\text{Cl}-\text{Cl}}-\text{B}_{\text{H}-\text{Cl}}$
$=218.5+122-433$
$=-92.5\text{kJ mol}^{-1}$
View full question & answer→Question 63 Marks
100mL of a liquid is contained in awn insulated container at a pressure of 1bar. The pressure is steeply increased to 100bar. The volume of the liquid is decreased by 1mL at this constant pressure. Find $\Delta\text{H}$ and $\Delta\text{U}.$
Answer$p_1=1$ bar, $p_2=100$ bar, $V_1=100 \mathrm{~mL}, V_2=99 \mathrm{~mL}$ For adiabatic process, $q=0$, $\Delta\text{U}=\text{q}+\text{W}$$\Delta\text{U}=\text{W}$
$\text{W}=-\text{p}\Delta\text{V}$
$=-100(99-100)=100\text{bar mL}$
$\Delta\text{H}=\Delta\text{U}+\Delta\text{pV}$
$=100+\text{p}_2\text{V}_2-\text{p}_1\text{V}_1$
$=100+(100\times99)-(1\times100)$
$=100+9900-100=9900\text{bar mL}$
View full question & answer→Question 73 Marks
Use the following data to calculate the $\Delta \mathrm{G}^{\circ}$ for the reaction in which $\mathrm{PF}_5(\mathrm{~g})$ forms from $\mathrm{PF}_3(\mathrm{~g})$ and $\mathrm{F}_2(\mathrm{~g})$ at 298 K .
$\mathrm{PF}_3(\mathrm{~g})+\mathrm{F}_2(\mathrm{~g}) \rightarrow \mathrm{PF}_5(\mathrm{~g})$
$\begin{matrix}\Delta \text{H}^{\circ}_{f} & -919 \text{kJ mol}^{-1} & 0 & -1577 \text{kJ} \text{ K}^{-1} \text{mol}^{-1} \ \\\text{S}^{\circ}& 273 & 203 & 301 \end{matrix}$
Answer$\Delta \text{H}^{\circ}= \sum\Delta \text{H}^{\circ}_f \text{(products)} - \sum\Delta \text{H}^{\circ}_f \text{(reactants)}$$= -1577 - (919)$
$= -658 \text{kJ}$
$\Delta \text{H}^{\circ}= \sum\text{S}^{\circ}_\text{products} - \sum \text{S}^{\circ}_\text{reactants}$
$= 301 - 273 - 203 = 175 \text{JK}^{-1}$
$= -0.175 \text{kJ} \text{ K}^{-1}$
$\Delta \text{G}^{\circ} = \Delta \text{H}^{\circ} - \text{T}\Delta \text{S}^{\circ}$
$= -658 \text{kJ} - 298 \times 0.175 \text{kJ}$
$= -606 \text{kJ}$
View full question & answer→Question 83 Marks
A man takes a diet equivalent to 10000kJ per day and does work, in expending his energy in all forms equivalent to 12500kJ per day. What is change in internal energy per day? If the energy lost was stored as sucrose (1632kJ per 100g), how many days should it take to lose 2kg of his weight? (Ignore water loss)
AnswerEnergy taken by a man = 10000kJ Change in internal energy per day = 12500 - 10000 = 2500kJ The energy is lost by the man as he expends more energy than he takes. Now 100g of sugar corresponds to energy = 1632kJ loss in energy. 2000g of sugar corresponds to energy $=\frac{1632\times2000}{100}$ = 32640kJ$\therefore$ Number of days required to lose 2000g of weight or 32640kJ of energy $=\frac{32640}{2500}=13\text{days}$
View full question & answer→Question 93 Marks
Given$\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \xrightarrow{ \ \ \ \ \ \ }2\text{NH}_3(\text{g}) ; \Delta_{\text{r}}\text{H}^\ominus= –92.4\text{kJ} \ \text{mol}^{–1}$
What is the standard enthalpy of formation of $\text{NH}_{3}$ gas?
AnswerStandard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state. Re-writing the given equation for 1 mole of $\text{NH}_{3(\text{g})}$,
$\frac{1}{2}\text{N}_{2(\text{g})} + \frac{3}{2}\text{H}_{2(\text{g})} \xrightarrow{ \ \ \ \ \ \ } \text{NH}_{3(\text{g})}$
$\therefore$ Standard enthalpy of formation of $\text{NH}_{3(\text{g})}$
$=\frac{1}{2}\Delta_{\text{r}}\text{H}^\ominus$
$=\frac{1}{2}(-92.4\text{kJ}) \ \text{mol}^{-1}$
$=-46.2\text{kJ} \ \text{mol}^{-1}$
View full question & answer→Question 103 Marks
Calculate the lattice enthalpy of $\mathrm{MgBr}_2$, given that
Enthalpy of formation of $\mathrm{MgBr}_2=-524 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Sublimation energy of $\mathrm{Mg}=148 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Ionization energy of $\mathrm{Mg}=2187 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Vaporisation energy of $\mathrm{Br}_2(\mathrm{l})=31 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Dissociation energy of $\mathrm{Br}_2(\mathrm{g})=193 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Electron gain enthalpy of $\mathrm{Br}_2(\mathrm{g})=331 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Answer$\Delta_\text{f}\text{H}=\Delta\text{H}_\text{sub}+\Delta\text{H}_\text{ion}+\Delta\text{H}_\text{EA}+\Delta\text{H}_\text{D}\\+\Delta\text{H}_\text{Lattice}+\Delta\text{H}_\text{vap}$$-524=148+2187-331+193\\+31+\Delta\text{H}_\text{Lattice}$
$\Delta\text{H}_\text{Lattice}=-524-2228=-2752\text{kJ mol}^{-1}$
View full question & answer→Question 113 Marks
Give the appropriate reason.
- It is preferable to determine the change in enthalpy rather than the change in internal energy.
- It is necessary to define the 'standard state'.
- It is necessary to specify the phases of the reactants and products in a thermochemical equation.
Answer
- Because it is easier to make measurement under constant pressure than under constant volume conditions.
- Enthalpy change depends upon the conditions in which a reaction is carried out. For making the comparison of results obtained by different people meaningful, the reaction conditions must be well-defined.
- Because enthalpy depends upon the phase of reactants and products.
View full question & answer→Question 123 Marks
Calculate the maximum work obtained when 0.75mol of an ideal gas expands isothermally and reversibly at 27°C from a volume of 15L to 25L.
Answer$\text{w}=-2.303\text{nRT}\log\frac{\text{V}_2}{\text{V}_1}$$\text{n}=0.75\text{mol},\ \text{R}=8.314\text{JK}^{-1}\text{mol}^{-1}\text{T}$
$\text{T}=27^\circ\text{C}+273=300\text{K}$
$\text{w}=-2.303\times0.75\times8.314\\\times300\text{K}\log\frac{25\text{L}}{15\text{L}}$
$\text{w}=-19.147\times\frac{3}{4}\times300[\log5-\log3]$
$\text{w}=\frac{-19.147\times900}{4}\times[0.6990-0.4771]$
$\text{w}=\frac{-19.147\times900\times0.2219}{4}$
$=\frac{-3823.84}{4}$
$=-955.96\text{J}$
View full question & answer→Question 133 Marks
Calculate the entropy change in surroundings when 1.00 mol of $H_2O(l)$ is formed under standard conditions. $ \Delta_{\text{f}}\text{H}^\ominus=-286\text{kJ} \ \text{mol}^{-1}.$
AnswerIt is given that $286 \mathrm{~kJ} \mathrm{~mol}^{-1}$ of heat is evolved on the formation of 1 mol of $\mathrm{H}_2 \mathrm{O}_{(l)}$. Thus, an equal amount of heat will be absorbed by the surroundings. $q_{\text {surr }}=+286 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Entropy change $\Delta\text{S}_{\text{surr}}$ for the surroundings $=\frac{\text{q}_{\text{surr}}}{7}$
$=\frac{286\text{kJ} \ \text{mol}^{-1}}{298\text{k}}$
$\therefore\Delta\text{S}_{\text{surr}}=959.73\text{J} \ \text{mol}^{-1} \ \text{k}^{-1}$
View full question & answer→Question 143 Marks
If the combustion of 1g of graphite produces 20.7kJ of heat, what will be molar enthalpy change? Give the significance of sign also.
AnswerHeat capacity of the calorimeter, Cv =20.7KJ/ mol The temperature change is not mentioned in the question, so we assume it to be, $\Delta\text{T}=1\text{K}$ Thus heat absorbed by the calorimeter = $\text{Cv}\times\Delta\text{T}$$=20.7\times1=20.7\text{KJ}$
This is the heat evolved in combustion of 1g graphite So, heat evolved in combustion of 1mole of graphite i.e 12g of graphite = 20.7 × 12 = 284.4KJ Since heat is evolved and vessel is closed enthalpy change of reaction = -284.4KJ/ mol −ve sign signifies heat evolved.
View full question & answer→Question 153 Marks
- Classify the following processes as reversible or irreversible:
- Dissolution of sodium chloride.
- Evaporation of water at 373K and 1atm pressure.
- Mixing of two gases by diffusion.
- Melting of ice without rise in temperature.
- When an ideal gas expands in vacuum, there is neither absorption nor evolution of heat. Why?
Answer
-
- Reversible.
- Irreversible.
- Irreversible.
- Reversible.
- It is because no work is done.
i.e., w = 0
$\because\text{w}=-\text{P}_\text{ext}\times\Delta\text{V}=0\times\Delta\text{V}=0$
$\Delta\text{U}=\text{q}+\text{w}$
q = 0 because gas chamber is insulated
$\therefore\Delta\text{U}=0+0=0$ View full question & answer→Question 163 Marks
For the reaction at 298K, 2A + B → C$\Delta\text{H} = 400\text{kJ} \ \text{mol}^{–1} \text{and} \Delta\text{S} = 0.2\text{kJ} \ \text{K}^{–1} \ \text{mol}^{–1} $
At what temperature will the reaction become spontaneous considering $\Delta\text{H}$ and $\Delta\text{S}$ to be constant over the temperature range.
AnswerFrom the expression,$\Delta\text{G} = \Delta\text{H} – \text{T}\Delta\text{S}$
Assuming the reaction at equilibrium, $\Delta\text{T}$ for the reaction would be:$\text{T} =( \Delta\text{H} –\Delta\text{G})\frac{1}{\Delta\text{S}}$
$=\frac{\Delta\text{H}}{\Delta\text{S}}$
$=\frac{400\text{kJ} \ \text{mol}^{-1}}{0.2\text{kJ} \ \text{mol}^{-1}}$
$\text{T} = 20000\text{K}$
For the reaction to be spontaneous, $\Delta\text{G}$ must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000K.
View full question & answer→Question 173 Marks
State the law of thermodynamics that was first formulated by Nernst in 1906. What is the utility of this law? The equilibrium constant for the reaction $A \rightleftharpoons B$ is $1.8 \times 10^{-7}$ at 298 K . Calculate the value of $\Delta G ^{\circ}$ for the reaction $\left( R =8.314 JK ^{-1} mol^{-1}\right)$. Predict the feasibility of the reaction under standard states.
AnswerAccording to Nernst, at absolute zero, the entropy of a perfectly crystalline substance is taken as zero at zero kelvin. This is also known as third law of thermodynamics. It helps us to calculate the absolute entropies of pure substances at different temperatures. $K =1.8 \times$ $10^{-7}, T=298 K, \Delta G ^{\circ}=?, R =8.314 JK ^{-1} mol^{-1} \Delta G ^{\circ}=-2.303 RT \log K$1$\Delta\text{G}^\circ=-2.303\text{RT}\log\text{K}$
$\Delta\text{G}^\circ=-2.303\times8.314\times298\log1.8\times10^{-7}$
$=-19.147\times298\log1.8\times10^{-7}$
$=-5705.8[\log1.8+\log10^{-7}]$
$=-5705.8[0.2552-7.0000]$
$=-5705.8\times(-6.7448)=+38484.47\text{J}$
$=38.484\text{kJ mol}^{-1}$
Since $\Delta\text{G}^\circ$ is +ve, therefore, reaction is not feasible at this temperature, i.e., reaction is non-spontaneous.
View full question & answer→Question 183 Marks
$6.72 \mathrm{dm}^3$ of an unknown gas at STP requires 113.35 J of heat to raise its temperature by $15^{\circ} \mathrm{C}$ at constant volume. Calculate $\mathrm{C}_{\mathrm{V}}, \mathrm{C}_{\mathrm{p}}$, and atomicity of the gas.
Answer$22.4 \mathrm{dm}^3$ of a gas at STP $=1 \mathrm{~mol} . \therefore 5.6 \mathrm{dm}^3$ of the gas at $\mathrm{STP}=\frac{1}{22.4} \times 6.72=0.3 \mathrm{~mol}$
Thus, for $15^{\circ}$ rise, 0.3 mol of the gas at constant volume require heat $=113.35 \mathrm{~J} . \therefore$ For $1^{\circ}$ rise, 1 mole of the gas at constant volume will require heat $=\frac{113.35}{15 \times 0.3} \mathrm{~J}=25.19 \mathrm{~J}$
$\therefore\text{C}_\text{V}=25.19\text{JK}^{-1}\text{mol}^{-1}$
Now, $\text{C}_\text{p}=\text{C}_\text{V}+\text{R}$
$=25.19\text{JK}^{-1}\text{mol}^{-1}+8.314\text{JK}^{-1}\text{mol}^{-1}$
$=33.5\text{JK}^{-1}\text{mol}^{-1};$
$\therefore\gamma=\frac{\text{C}_\text{p}}{\text{C}_\text{V}}=\frac{33.5}{25.19}=1.33$
View full question & answer→Question 193 Marks
1 mole of an ideal gas undergoes reversible isothermal expansion from an initial volume of $V _1$ to a final volume of $10 V_1$ and does 10 kJ of work. The initial pressure was $1 \times 10^7 Pa$.
i. Calculate $V _1$.
ii. If there were 2 moles of gas, what must its temperature have been?
Answer
- We know that $\text{W}=-2.303\text{nRT}\log\frac{\text{V}_2}{\text{V}_1}$
$10\times10^3\text{J}=2.303\times1\times8.314\\\times\text{T}\times\log\frac{10\text{V}_1}{\text{V}_1}$
T = 522.3K
For initial conditions, $p_1V_1 = n_1RT,$
$i.e., (10^7Pa)V_1 = 1 \times 8.314 \times 522.3$
$V_1 = 4.342 \times 10^{-4}m^3$
$= 4.342 \times 10^2cm^3$
$= 434.2cm^3$We cannot apply the formula $- W = p \Delta V$ because expansion is not against constant pressure.
iii. If there were 2 moles of the gas, applying $p_1 V_1=n_1 R T$, we get $\left(10^7 Pa\right)\left(4.342 \times 10^{-4} m^3\right)=2 \times 8.314 T$ or $T =$ 261.1 K , i.e. half of the first value. View full question & answer→Question 203 Marks
- State first law of thermodynamics.
- Heat (q) and work function (w) individually are not state functions but their sum is always a state function. Why?
- Define extensive property with example.
Answer
- First law of thermodynamics: It states energy can neither be created, nor be destroyed. It can change from one from to another. The total energy of universe remains constant.
$\Delta\text{U}=\text{q}+\text{w}$
- 'q' and 'w' depend upon path, therefore, these are path function.
$\text{q}+\text{w}=\Delta\text{U}$
$\Delta\text{U}$ is internal energy change which is state function because it depends upon initial and final state of the system and not on path, therefore, 'q + w' is state function.
- Extensive property: The property which depends upon amount of substance is called extensive property, e.g., mass, volume are extensive properties.
View full question & answer→Question 213 Marks
Starting with the thermodynamic relationship G = H - TS, derive the following relationship: $\Delta\text{G}=-\text{T}\Delta\text{S}_\text{Total}$
Answer$\text{G}=\text{H}-\text{TS},$ $\text{G}_1=\text{H}_1-\text{TS}_1,$ $\text{G}_2=\text{H}_2-\text{TS}_2$$\text{G}_2-\text{G}_1=\text{H}_2-\text{H}_1-\text{T}(\text{S}_2-\text{S}_1)$
$\Delta\text{G}=\text{H}_2-\text{H}_1-\text{T}(\text{S}_2-\text{S}_1)$
$\Delta\text{G}=\Delta\text{H}-\text{T}\Delta\text{S}$
$\Delta\text{S}_\text{Total}=\Delta\text{S}_\text{sys}+\Delta\text{S}_\text{surr}$
$\Rightarrow\Delta\text{S}_\text{Total}=\Delta\text{S}_\text{sys}-\frac{\text{q}}{\text{T}}$
$\Delta\text{S}_\text{sys}=\frac{+\text{q}}{\text{T}},\Delta\text{S}_\text{surr}=\frac{-\text{q}}{\text{T}}$
$\Rightarrow\text{T}\Delta\text{S}_\text{Total}=\text{T}\Delta\text{S}_\text{sys}-\text{q}=\text{T}\Delta\text{S}_\text{sys}-\Delta\text{H}$
$\text{T}\Delta\text{S}_\text{Total}=-\Delta\text{G}$
$\Rightarrow\Delta\text{G}=-\text{T}\Delta\text{S}_\text{Total}$
View full question & answer→Question 223 Marks
Why is the entropy of a substance taken as zero at 0K? Calculate the standard Gibbs free energy change for the reaction:$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g})$ at 298K
The value of equilibrium constant for the above reaction is $6.6 \times 10^5 \cdot\left[R=8.314 \mathrm{J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$
AnswerThe entropy of all substances at absolute zero (0K) is taken as zero because of complete order in the system, i.e. the atoms or molecule do not move at all in the perfectly crystalline state. $\Delta\text{G}^\circ=-2.303\text{RT}\log\text{K}$
$=-2.303\times8.314\text{JK}^{-1}\text{mol}^{-1}\\\times298\text{K}\log6.6\times10^5$
$=-19.147\text{J}\times298\log6.6\times10^5$
$=-5705.8[\log6.6+\log10^5]$
$=-5705.8[0.8195+5.0000]$
$=-5705.8\times5.8195\text{J}=-33204.903\text{J}$
$\Delta\text{G}^\circ=-33.205\text{kJ mol}^{-1}$
View full question & answer→Question 233 Marks
Calculate the standard Gibbs energy change for the formation of propane at 298K.$3\text{C(graphite)}+4\text{H}_2\text{O}\overrightarrow{\ \ \ \ \ \ \ }\ \text{C}_3\text{H}_8(\text{g})$
$\Delta_\text{f}\text{H}^\circ$ for propane, $\mathrm{C}_3 \mathrm{H}_8(\mathrm{~g})$ is $-103.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Given: $\text{S}^\circ_\text{m}\text{C}_3\text{H}_8(\text{g})=270.2\text{JK}^{-1}\text{mol}^{-1},$ $\text{S}^\circ_\text{m}\text{C}_{(\text{graphite})}=5.70\text{JK}^{-1}\text{mol}^{-1}$ and $\text{S}^\circ_\text{m}\text{H}_2(\text{g})=130.7\text{JK}^{-1}\text{mol}^{-1}$
Answer$3\text{C}_\text{(graphite)}+4\text{H}_2\text{O}\overrightarrow{\ \ \ \ \ \ \ }\ \text{C}_3\text{H}_8(\text{g})$$\Delta_\text{r}\text{S}^\circ=\sum\text{S}^\circ_\text{m}(\text{products})-\text{S}^\circ_\text{m}(\text{reactants})$
$\Delta_\text{r}\text{S}^\circ=\text{S}^\circ_\text{m}[\text{C}_3\text{H}_8(\text{g})]-3\text{S}^\circ_\text{m}[\text{C(graphite)}]\\-4\text{S}^\circ_\text{m}[\text{H}_2]$
$=(270.2-3\times5.70-4\times130.7)\text{JK}^{-1}\text{mol}^{-1}$
$=(270.2-17.10-522.80)\text{JK}^{-1}\text{mol}^{-1}$
$=(270.2-539.90)=-269.7\text{JK}^{-1}\text{mol}^{-1}$
$\Delta_\text{r}\text{H}^\circ=\Delta_\text{f}\text{H}^\circ_\text{products}-\Delta_\text{f}\text{H}^\circ_\text{reactants}$
$\Delta_\text{r}\text{H}^\circ=\Delta_\text{f}\text{H}^\circ(\text{C}_3\text{H}_8)-4\Delta_\text{f}\text{H}^\circ[\text{H}_2(\text{g})]\\-3\Delta_\text{f}\text{H}^\circ[\text{C(graphipte)}]$
$\Delta_\text{r}\text{H}^\circ=-103.8\text{kJ mol}^{-1}$
$\Delta_\text{r}\text{G}^\circ=\Delta_\text{r}\text{H}^\circ-\text{T}\Delta_\text{r}\text{S}^\circ$
$=-103.8-\Big(\frac{298\times(-269.7)}{1000}\Big)\text{kJ mol}^{-1}$
$=(-103.80+80.370)\text{kJ mol}^{-1}$
$=-23.43\text{kJ mol}^{-1}$
View full question & answer→Question 243 Marks
Comment on the thermodynamic stability of NO(g), given$\frac{1}{2}\text{N}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g})\xrightarrow{ \ \ \ \ \ \ } \text{NO}_{(\text{g})} \ ; \ \Delta_{\text{r}}\text{H}^\ominus=90\text{kJ} \ \text{mol}^{-1}$
$\text{NO}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g})\xrightarrow{ \ \ \ \ \ \ } \text{NO}_2(\text{g}) \ ; \ \Delta_{\text{r}}\text{H}^\ominus=-74\text{kJ} \ \text{mol}^{-1}$
AnswerThe positive value of $\Delta_{\mathrm{r}} \mathrm{H}$ indicates that heat is absorbed during the formation of $\mathrm{NO}_{(\mathrm{g})}$. This means that $\mathrm{NO}_{(\mathrm{g})}$ has higher energy than the reactants ( $\mathrm{N}_2$ and $\mathrm{O}_2$ ). Hence, $\mathrm{NO}_{(\mathrm{g})}$ is unstable.
The negative value of $\Delta_{\mathrm{r}} \mathrm{H}$ indicates that heat is evolved during the formation of $\mathrm{NO}_{2(\mathrm{~g})}$ from $\mathrm{NO}_{(\mathrm{g})}$ and $\mathrm{O}_{2(\mathrm{~g})}$. The product, $\mathrm{NO}_{2(g)}$ is stabilized with minimum energy.
Hence, unstable $\mathrm{NO}_{(\mathrm{g})}$ changes to stable $\mathrm{NO}_{2(g)}$.
View full question & answer→Question 253 Marks
How is the third law of thermodynamics useful in calculation of the absolute entropies? Calculate the value of $\Delta S ^{\circ}$ for the following reaction at 400 K :
$2 NOCl(g) \rightarrow 2 NO(g)+Cl_2(g)$
If the value of equilibrium constant for the reaction at 400 K is $1.958 \times 10^{-4}$ and $\Delta H ^{\circ}=77.2 kJ mol ^{-1}\left( R =8.314 J K ^{-1} mol^{-1}\right)$.
Answer$\Delta S = S _{ T }- S _0$ Where $S _{ T }$ is entropy at temperature TK
But from third law, $S _0=0$
Where $S _0$ is entropy at 0 K
$\therefore S_{T}=\Delta S$
If we measure the energy required to raise the temperature of a crystalline substance from 0 K to 298 K , we can determine entropy change. The entropy $S _{ T }$ at 298 K is called absolute entropy.
$\Delta\text{G}^\circ=-2.303\text{RT}\log\text{K}$
$=-2.303\times8.314\text{JK}^{-1}\text{mol}^{-1}\\\times400\text{K}\log1.958\times10^{-4}$
$=(-19.147\times400\log1.958\times10^{-4})\text{J mol}^{-1}$
$=[-7658.8\times(\log1.958+\log10^{-4})]\text{J mol}^{-1}$
$=[-7658.8\times(0.2917-4.0000)]\text{J mol}^{-1}$
$\Delta\text{G}^\circ=-7658.8\times-3.7183=+28477.72\text{J mol}^{-1}$
$=28.48\text{kJ mol}^{-1}$
$\Delta\text{G}^\circ=\Delta\text{H}^\circ-\text{T}\Delta\text{S}^\circ$
$28.48\text{kJ mol}^{-1}=77.20\text{kJ mol}^{-1}-400\Delta\text{S}^\circ$
$-400\Delta\text{S}^\circ=-48.73\text{kJ mol}^{-1}$
$\Delta\text{S}^\circ=\frac{48.73\times1000\text{J mol}^{-1}}{400\text{K}}=\frac{487.3}{4}$
$=121.82\text{JK}^{-1}\text{mol}^{-1}$
View full question & answer→Question 263 Marks
Explain the term 'standard molar free energy of formation' of a compound. Calculate the equilibrium constant for the reaction: $2 SO _2(g)+ O _2(g) \rightleftharpoons 2 SO _3(g)$ at $25^{\circ} C$.
Given: $\Delta_{ f } G ^{\circ} SO _3(g)=-371.1 kJ mol { }^{-1}, \Delta_{ f } G ^{\circ} SO _2(g)=-300.2 kJ mol { }^{-1}, R =8.31 JK ^{-1} mol^{-1}$.
AnswerStandard free energy of formation of compound is defined as free energy change when a compound is formed from its constituting elements at 298 K and 1atm pressure. $2 SO _2(g)+ O _2(g) \rightleftharpoons 2 SO _3(g)$
$\Delta\text{G}^\circ=\sum\Delta_\text{f}\text{G}^\circ(\text{products})-\sum\Delta_\text{f}\text{G}^\circ(\text{reactants})$
$=2\Delta_\text{f}\text{G}^\circ(\text{SO}_3)-2\Delta_\text{f}\text{G}^\circ(\text{SO}_2)-\Delta_\text{f}\text{G}^\circ(\text{O}_2)$
$=2\times(-371.1)-2\times(-300.2)-0$
$=-742.2+600.4=-141.8$
$\Delta\text{G}^\circ=-2.303\text{RT}\log\text{K}$
$-141.8\times1000\text{J}=-2.303\times8.314\times298\log\text{K}$
$\log\text{K}=\frac{1,41,800}{5705.8}=24.8519$
$\Rightarrow\text{K}=7.114\times10^{24}$
[Antilog of $0.8519 = 7.114]$
[Antilog of $24.8519 = 7.114 \times 10^{24}]$
View full question & answer→Question 273 Marks
Calculate the equilibrium constant for the following reaction at 298K and 1atm pressure $\text{NO(g)}+\frac{1}{2}\text{O}_2(\text{g})\rightleftharpoons\text{NO}_2(\text{g})$
Given $\Delta \mathrm{H}_{\mathrm{f}}^{\circ}$ at $298 \mathrm{~K}^{-20} \mathrm{NO}(\mathrm{g})=90.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\mathrm{NO}_2(\mathrm{~g})=33.8 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta \mathrm{~S}^{\circ}$ at $298 \mathrm{~K}^{-}$for the reaction $=-70.0 \mathrm{JK}^{-}$ ${ }^1 \mathrm{~mol}^{-1}, \mathrm{R}=8.31 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$.
AnswerEnthalpy change for the reaction, $\Delta\text{H}^\circ=\sum\Delta_\text{f}\text{H}^\circ_{(\text{products})}-\sum\Delta_\text{f}\text{H}^\circ_{(\text{reactants})}$$\big[\Delta_\text{f}\text{H}^\circ\text{NO}_2(\text{g})\big]-\Big[\Delta_\text{f}\text{H}^\circ\text{NO(g)}+\frac{1}{2}\Delta_\text{f}\text{H}^\circ\text{O}_2(\text{g})\Big]$
$=[33.8]-\Big[90.4+\frac{1}{2}\times0\Big]$
$=-56.6\text{kJ mol}^{-1}$
$\Delta\text{S}^\circ=-70\text{JK}^{-1}\text{mol}^{-1}$
Now, $\Delta\text{G}^\circ=\Delta\text{H}^\circ-\text{T}\Delta\text{S}^\circ,\ \text{T}=298\text{K}$$\therefore\Delta\text{G}^\circ=-56600-298\times(-70)$
$=-35740\text{J mol}^{-1}$
$\log\text{K}=-\frac{\Delta\text{G}}{2.303\text{RT}}$
$\log\text{K}=-\frac{-35740}{2.303\times8.31\times298}=6.267$
$\therefore\text{K}=\text{anti}\log(6.267)=1.85\times10^6$
View full question & answer→Question 283 Marks
How will you calculate work done on an ideal gas in a compression, when change in pressure is carried out in infinite steps?
Answer
The work done can be calculated with the help of p–V plot. A p–V plot of the work of compression which is carried out by change in pressure in infinite steps, is given in Fig. Shaded area represents the work done on the gas.
View full question & answer→Question 293 Marks
- State Hess's Law of constant heat summation. How does it follow from the first law of thermodynamics.
- Determine $\Delta_\text{r}\text{H}^\circ,$ at 298K for reaction:
$\text{C(graphite)}+2\text{H}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ \ }\ \text{CH}_4(\text{g});\ \Delta_\text{r}\text{H}^\circ=?$
You are given:
- $\text{C(graphite)}+\text{O}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ }\ \text{CO}_2(\text{g});$ $\Delta_\text{r}\text{H}^\circ=-393.51\text{kJ/ mol}$
- $\text{H}_2\text{(g)}+\frac{1}{2}\text{O}_2(\text{g})\overrightarrow{ \ \ \ \ \ \ \ }\ \text{H}_2\text{O(l)};$ $\Delta_\text{r}\text{H}^\circ=-285.8\text{kJ/ mol}$
- $\text{CO}_2(\text{g})+2\text{H}_2\text{O(l)}\overrightarrow{\ \ \ \ \ \ \ \ \ }\ \text{CH}_4(\text{g})+2\text{O}_2(\text{g});$ $\Delta_\text{r}\text{H}=+890.3\text{kJ/ mol}$
Answer
- Hess's law states enthalpy change remains the same whether the reaction takes place in one step or in several steps.
It follows from Ist law of thermodynamics that energy can neither be created nor be destroyed. It can change from one form to another. The total energy of universe remains constant.
- $\text{CO}_2(\text{g})+2\text{H}_2\text{O(l)}\overrightarrow{ \ \ \ \ \ \ }\ \text{CH}_4(\text{g})+2\text{O}_2(\text{g})$
$\Delta_\text{r}\text{H}=\Delta_\text{f}\text{H}(\text{CH}_4)+2\Delta_\text{f}\text{H}(\text{O}_2)-\Delta_\text{f}\text{H}(\text{CO}_2)\\-2\Delta_\text{f}\text{H}[\text{H}_2\text{O(l)}]$
$+890.3\text{kJ mol}^{-1}=\Delta_\text{f}\text{HCH}_4+2\times0-\\-(-393.51\text{kJ mol}^{-1})-2\times(-285.8\text{kJ mol}^{-1})$
$\Delta_\text{f}\text{H}(\text{CH}_4)=(+890.3-393.51\text{kJ}\\-571.6)\text{kJ mol}^{-1}$
$\Delta_\text{f}\text{H}(\text{CH}_4)=-74.8\text{kJ mol}^{-1}$ View full question & answer→Question 303 Marks
For the reaction,
$2 \mathrm{Cl}(\mathrm{g}) \rightarrow \mathrm{Cl}_2(\mathrm{g})$, what are the signs of $\Delta\text{H}$ and $\Delta\text{S}?$
Answer$\Delta\text{H}$ and $\Delta\text{S}$ are negativeThe given reaction represents the formation of chlorine molecule from chlorine atoms. Here, bond formation is taking place. Therefore, energy is being released. Hence, $\Delta\text{H}$ is negative.
Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, $\Delta\text{S}$ is negative for the given reaction.
View full question & answer→Question 313 Marks
QUESTION Use the following data to calculate $\Delta_{\text {lattice }} H ^{\ominus}$ for $NaBr \Delta_{\text {sub }} H ^{\ominus}$ for sodium metal $=108.4 kJ mol ^{-1}$ Ionization enthalpy of sodium $=496 kJ mol ^{-1}$ Electron gain enthalpy of bromine $=-325 kJ mol ^{-1}$ Bond dissociation enthalpy of bromine $=192 kJ$ $mol ^{-1} \Delta_{ f } H ^{\ominus}$ for $NaBr ( s )=-360.1 kJ mol ^{-1}$
Answer$\text{Na}(\text{s})\rightarrow\text{Na}(\text{g}),\Delta_\text{sub}\text{H}^\circ=108.4\text{kJ }\text{mol}^{-1}$$\text{Na}\rightarrow\text{Na}^{+}+\text{e}^-,\Delta_\text{i}\text{H}^\circ=496\text{kJ }\text{mol}^{-1}$
$\frac{1}{2}\text{Br}_2\rightarrow\text{Br},\frac{1}{2}\Delta_\text{diss}\text{H}^\circ=\frac{192}{2}=96\text{kJ }\text{mol}^{-1}$
$\text{Br}+\text{e}^-\rightarrow\text{Br}^-,\Delta_\text{eg}\text{H}^\circ=-325\text{kJ }\text{mol}^{-1}$
$\Delta_\text{f}\text{H}^\circ\ \text{for}\ \text{NaBr}=-360.1\text{kJ }\text{mol}^{-1}$
Enthalpies in diffrent steps involved in the formation of NaBr(s) are,
$\Delta_\text{f}\text{H}^\circ=\Delta_\text{sub}\text{H}^\circ+\frac{1}{2}\Delta\text{diss}\text{H}^\circ+\Delta_\text{i}\text{H}^\circ+\Delta_\text{eg}\text{H}^\circ+\Delta_\text{lattice}\text{H}^\circ$
$\therefore\ \Delta_\text{lattice}\text{H}^\circ=-360.1-108.4-96-496+325$
$=-735.5\text{kJ }\text{mol}^{-1}$
View full question & answer→Question 323 Marks
Enthalpy of combustion of carbon to $CO _2$ is $-393.5 kJ mol ^{-1}$. Calculate the heat released upon formation of 35.2 g of $CO _2$ from carbon and dioxygen gas.
AnswerFormation of $CO _2$ from carbon and dioxygen gas can be represented as: $C _{( s )}+ O _{2(g)} \longrightarrow CO _{2(g)} \Delta_{ f } H =-393.5 kJ mol ^{-1}$ Heat released on formation of $44 g CO _2=-393.5 kJ mol { }^{-1}$
Heat released on formation of $35.2 g CO _2$
$=\frac{-393.5\text{kJ} \ \text{mol}^{-1}}{44\text{g}}\times35.2\text{g}$
$= –314.8\text{kJ} \ \text{mol}^{–1} $
View full question & answer→Question 333 Marks
QUESTION Calculate bond energy of C - H bond if $\Delta_{ c } H$ of $CH _4$, is $-891 kJ mol { }^{-1}, \Delta_c H$ of $C ( s )$ is $-394 kJ mol ^{-1}, \Delta_c H ^{-}$of $H _2(g)$ is $-286 kJ Jol ^{-1}$, heat of sublimation of $C ( s )$ is $717 kJ mol ^{-1}$, heat of dissociation of $H _2$ is $436 kJ mol ^{-1}$.
Answer$\text{CH}_4(\text{g})+2\text{O}_2(\text{g})\overrightarrow{\ \ \ \ \ \ }\ \text{CO}_2(\text{g})+2\text{H}_2\text{O(l)}$$\Delta\text{H}=-891\text{kJ mol}^{-1}\dots(\text{i})$
$\text{C(s)}+\text{O}_2(\text{g})\overrightarrow{\ \ \ \ \ \ }\ \text{CO}_2(\text{g})$
$\Delta\text{H}=-394\text{kJ mol}^{-1}\dots(\text{ii})$
$\text{H}_2(\text{g})+\frac{1}{2}\text{O}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ }\ \text{H}_2\text{O(l)}$
$\Delta\text{H}=-286\text{kJ mol}^{-1}\dots(\text{iii})$
$\text{C(s)}\overrightarrow{\ \ \ \ \ \ }\ \text{C(g)}$
$\Delta\text{H}=+717\text{kJ mol}^{-1}\dots(\text{iv})$
$\text{H}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ }\ 2\text{H(g)}$
$\Delta\text{H}=+436\text{kJ mol}^{-1}\dots(\text{v})$
Target equation in $\ \ \ \ \ \ \ \ \text{H}\\ \ \ \ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{H}\overrightarrow{\ \ \ \ \ \ \ \ }\ \text{C(g)}+4\text{H(g)}\\ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \text{H}$
Reversing (ii), Multiplying (iii) by 2 and reversing and Multiplying (v) by 2 and then adding all together we get:
$\text{CH}_4\overrightarrow{\ \ \ \ \ }\ \text{C(g)}+4\text{H(g)}$
$\Delta\text{H}=-891+394+572+717+872$
$\Delta\text{H}=1664\text{kJ}$
Now, energy required to break 4 C-H bond = 1664kJ.
Therefore, energy required to break 1 C-H bond $=\frac{1664}{4}=416\text{kJ mol}^{-1}.$
View full question & answer→Question 343 Marks
Reaction $\text{X} → \text{Y}\Delta\text{H}= \text{+ve}$ is spontaneous at temperature ‘T'. Determine:
- Sign of AS for this reaction.
- Sign of AG for ‘Y' to 'X'
- Sign of AG at temperature < T.
Answer
- $\Delta \text{S} = +\text{ve}$ for this reaction
$\because\Delta \text{G} = -\text{ve} \text{ only if} \ \Delta \text{S} $
$= + \text{ve} \text{ as }\Delta \text{H} = +\text{ve}$
- $\Delta \text{G} = +\text{ve} \text{ for } '\text{y}' \text{ to } '\text{x}' $
because $ \Delta\text{G} = - \text{ve}$ from
'X' to 'Y' as it is spontaneous
- $\Delta \text{G}^{\circ} = \Delta \text{H}^{\circ} - \text{T}\Delta \text{S}^{\circ}$
Since $\Delta \text{H}^{\circ} = + \text{ve}, \Delta \text{S} = + \text{ve}$
At temp T, $\Delta \text{G} = -\text{ve,}$ reaction is spontaneous Below temp T, $\Delta \text{G} = +\text{ve,}$reaction will be non-spontaneous. View full question & answer→Question 353 Marks
$\text{X(g)}+3\text{Y(g)}\rightleftharpoons2\text{Z(g)};$ $\Delta\text{H}=-40\text{kJ}$ and $\text{s}^\circ$ of X, Y and Z are 60, 40 and $50 \mathrm{~kJ}^{-1} \mathrm{~mol}^{-1}$ respectively. Calculate the temperature above or below which reaction become feasible?
Answer$\Delta\text{S}^\circ=\sum\text{S}^\circ_\text{products}-\sum\text{S}^\circ_\text{reactants}$$=2\times50-60-3\times40$
$=100-60-120$
$=-80\text{Jk}^{-1}\text{mol}^{-1}$
$\Delta\text{G}^\circ=\Delta\text{H}^\circ-\text{T}\Delta\text{S}$
$0=-40000\text{J}-\text{T}\times(-80)$
$80\text{T}=40000$
$\text{T}=500\text{K}$
Below 500K, reaction will be spontaneous.
$\therefore\Delta\text{G}$ will be -ve because both $\Delta\text{H}$ and $\Delta\text{S}$ are - ve.
$\therefore\Delta\text{H}>\text{T}\Delta\text{S}$
View full question & answer→Question 363 Marks
What is the value of equilibrium constant for the following reaction at 400K?$2\text{NOCl(g)}\rightleftharpoons2\text{NO(g)}+\text{Cl}_2(\text{g})$
$\Delta\text{H}^\circ=77.5\text{kJ mol}^{-1},$ $\text{R}=8.314\text{J mol}^{-1}\text{K}^{-1},$ $\Delta\text{S}=135\text{J K}^{-1}\text{mol}^{-1}.$
Answer$\Delta\text{G}=\Delta\text{H}-\text{T}\Delta\text{S}$$\Delta\text{G}=77.5\times1000\text{ J}-400\text{K}\times135\text{J K}^{-1}$
$=77500\text{J}-54000\text{J}=23500\text{J}$
$\Delta\text{G}=-2.303\text{RT}\log\text{K}$
$23500\text{J}=-2.303\times8.314\times400\text{K}\log\text{K}$
$\log\text{K}=\frac{-23500}{19.147\times400\text{K}}=\frac{-235}{76.588}$
$=-3.068+1-1=\bar{4}.932$
$\text{K}=\bar{4}.932,\text{ Antilog}=8.551\times10^{-4}$
$\text{Antilog }0.932=8.551$
$\text{Antilog }\bar{4}.932=8.551\times10^{-4}$
View full question & answer→Question 373 Marks
10g of argon is compressed isothermally and reversibly at a temperature of 27°C from 10L to 5L. Calculate q, W, $\Delta\text{U}$ and $\Delta\text{H}$ for this process. $R = 2.0cal K^{-1} mol^{-1}$, log2 = 0.30, atomic weight, of Ar = 40.
Answer$\text{q}=2.303\text{nRT}\log\frac{\text{V}_2}{\text{V}_1}$$=2.303\times\frac{10}{40}\times300\times\log\frac{5}{10}$
$=-103.635\text{cal}$
For isothermal expansion, $\Delta\text{U}=0$
$\text{W}=\Delta\text{U}-\text{q}=0-(-103.635)$
$=+103.635\text{cal}$
Also, when temperature is constant,
$p_1V_1 = p_2V_2$ or $pV$ = constant
$\Delta\text{H}=\Delta\text{E}+\Delta(\text{pV})=0+0=0$
View full question & answer→Question 383 Marks
QUESTION Calculate the standard Gibbs energy change for the formation of propane at $298 K .3 C ( graphite )+4 H _2(g) \rightarrow C _3 H _8(g) \Delta_{ f } H ^{\circ}$ for propane, $C _3 H _8(g)$ is $-103.8 kJ mol ^{-1}$
Given, $S _{ m }^{\circ} C _3 H _8(g)=270.2 JK ^{-1} mol^{-1} S_{ m }^{\circ} C ($ graphite $)=5.70 JK ^{-1} mol^{-1}$ and $S _{ m }^{\circ} H _2(g)=130.7 JK ^{-1} mol^{-1}$
Answer3C(graphite) $+ 4H_2(g) → C_3H_8$(g)$\Delta_\text{r}\text{S}=\sum\text{S}^\circ_\text{m}(\text{products})-\text{S}^\circ_\text{m}(\text{reactants})$
$\Delta_\text{r}\text{S}=\text{S}^\circ_\text{m}[\text{C}_3\text{H}_8(\text{g})-3\text{S}^\circ_\text{m}[\text{C(graphite)}]\\+4\text{S}^\circ_\text{m}[\text{H}_2]]$
$=(270.2-3\times5.70-4\times130.7)\text{JK}^{-1}\text{mol}^{-1}$
$=(270.2-17.10-522.80)\text{JK}^{-1}\text{mol}^{-1}$
$=(270.2-539.90)=-269.7\text{JK}^{-1}\text{mol}^{-1}$
$\Delta_\text{r}\text{H}^\circ=\Delta_\text{f}\text{H}^\circ\text{products}-\Delta_\text{f}\text{H}^\circ\text{reactants}$
$\Delta_\text{r}\text{H}^\circ=\Delta_\text{f}\text{H}^\circ(\text{C}_3\text{H}_8)-4\Delta_\text{f}\text{H}^\circ[\text{H}_2(\text{g})]\\-3\Delta_\text{f}\text{H}^\circ[\text{C(graphite)}]$
$\Delta_\text{r}\text{H}^\circ=-103.8\text{kJ mol}^{-1};\ \Delta_\text{r}\text{G}^\circ=\Delta_\text{r}\text{H}^\circ-\text{T}\Delta_\text{r}\text{S}^\circ$
$=\Big(-103.8-\frac{298\times(-269.7)}{1000}\Big)\text{kJ mol}^{-1}$
$=(-103.80+80.370)\text{kJ mol}^{-1}$
$=-23.43\text{kJ mol}^{-1}$
View full question & answer→Question 393 Marks
What is meant by entropy? Predict the sign of entropy change in each of the following:
- $\left.\mathrm{H}_2 \text { (at } 298 \mathrm{~K}, 1 \mathrm{~atm}\right) \rightarrow \mathrm{H}_2 \text { (at } 298 \mathrm{~K}, 10 \text { atm) }$
- $\mathrm{H}_2 \mathrm{O}(\text { at } 298 \mathrm{~K}, 1 \mathrm{~atm}) \rightarrow \mathrm{H}_2 \mathrm{O} \text { (at } 330 \mathrm{~K}, 1 \text { atm) }$
- $2 \mathrm{NH}_4 \mathrm{NO}_3(\mathrm{~s}) \rightarrow 2 \mathrm{~N}_2(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{O}_2(8)$
- Crystallization of copper sulphate from its saturated solution.
- $2\text{SO}_2(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{SO}_3(\text{g}).$
AnswerEntropy is defined as the degree of randomness or disorder.
- $\Delta\text{S}=-\text{ve}$
- $\Delta\text{S}=+\text{ve}$
- $\Delta\text{S}=+\text{ve}$
- $\Delta\text{S}=-\text{ve}$
- $\Delta\text{S}=-\text{ve}$
View full question & answer→Question 403 Marks
Choose the correct answer.
A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be:
- Possible at high temperature.
- Possible only at low temperature.
- Not possible at any temperature.
- Possible at any temperature.
Answer
- Possible at any temperature
Explanation:
For a reaction to be spontaneous, $\Delta\text{G} $ should be negative.
$\Delta\text{G} = \Delta\text{H} – \text{T}\Delta\text{S}$
According to the question, for the given reaction,
$\Delta\text{S} = \text{positive} $
$\Delta\text{H} = \text{negative}$ (since heat is evolved)
$\Delta\text{G} = \text{negative}$
Therefore, the reaction is spontaneous at any temperature. View full question & answer→Question 413 Marks
Calculate the work done when$ 11.2g$ of iron dissolves in hydrochloric acid in:
- A closed vessel.
- An open beaker at $25^\circ{C}$ (Atomic mass of Fe$ = 56u).$
AnswerIron reacts with hydrochloric acid to produce H , gas as $Fe ( s )+2 HCl ( aq ) \rightarrow FeCl _2( aq )+ H _2(g) Thus$, 1 mole of Fe , i.e. 56 g Fe produces $H _2$ gas $=1 mol . \therefore 11.2 g$ Fe will produce $H _2$ gas $=\frac{1}{56} \times 11.2=0.2 mol$
i. If the reaction is carried out in closed vessel, $\Delta V =0$
$\therefore W=-P_{ext} \Delta V=0$
ii. If the reaction is carried out in open beaker (external pressure being 1atm)
Initial volume $=0$ (because no gas is present)
Final volume occupied by 0.2 mole of $H _2$ at $25^{\circ} C$ and 1atm pressure can be calculated as follows $pV = nRT$
$\therefore\text{V}=\frac{\text{nRT}}{\text{p}}$
$=\frac{0.2\text{mol}\times0.0821\text{L atm K}^{-1}\text{mol}^{-1}\times298\text{K}}{1\text{atm}}$
$=4.89\text{L}$
$\therefore\Delta\text{V}=\text{V}_\text{final}-\text{V}_\text{initial}=4.89\text{L}$
$\text{W}=-\text{p}_\text{ext}\Delta\text{V}=-1\text{atm}\times4.89\text{L}$
$=-4.89\text{L atm}=-4.89\times101.3\text{J}$
$=-495.4\text{J}$
View full question & answer→MCQ 423 Marks
Choose the correct answer.$\Delta\text{U}^\ominus$of combustion of methane is $-\mathrm{X} \mathrm{~kJ} \mathrm{~mol}^{-1}$. The value of $\Delta\text{H}^\ominus$ is:
- A
$=\Delta\text{U}^\ominus$
- B
$>\Delta\text{U}^\ominus$
- ✓
$<\Delta\text{U}^\ominus$
- D
$=0$
AnswerCorrect option: C. $<\Delta\text{U}^\ominus$
- $<\Delta\text{U}^\ominus$
Explanation:
The balanced chemical equation fir the combustion reaction is:
$\text{CH}_{4(\text{g})}+2\text{O}_{{2(\text{g})}\xrightarrow[]{\ \ \ \ \ \ \ \ \ \ \ \ \ }\text{CO}_2(\text{g})}+2\text{H}_2\text{O}(\text{l})$
$\Delta_{\text{ng}}=1-3=-2$
$\Delta\text{H}^\ominus=\Delta\text{U}^\ominus+\Delta_\text{ng}\text{RT}=\Delta\text{U}^\ominus-2\text{RT}$
$\therefore \Delta\text{H}^\ominus<\Delta\text{U}^\ominus$ View full question & answer→Question 433 Marks
10 moles of an ideal gas expand isothermally and reversibly from a pressure of 5atm to 1atm at 300K. What is the largest mass that can be lifted through a height of 1m by this expansion?
Answer$\text{W}_\text{exp}=-2.303\text{nRT}\log\frac{\text{p}_1}{\text{p}_2}$$=-2.303(10)\times(8.314)(300)\log\frac{5}{1}$
$=-40.15\times10^3\text{J}$
If M is the mass that can be lifted by this work through a height of 1m, then work done = Mgh
$40.15\times10^3\text{J}=\text{M}\times9.81\times\text{ms}^{-1}\times1\text{m}$
$\text{M}=\frac{40.15\times10^3\text{kg m}^2\text{s}^{-2}}{9.81\text{m s}^{-2}\times1\text{m}}$ $\big(\because\text{J}=\text{kg m}^2\text{s}^{-2}\big)$
$=4092.76\text{kg}$
View full question & answer→Question 443 Marks
Using the data (all values are in kilocalories per mole at $25^{\circ} C$ ) given below, calculate the bond energy of $C - C$ and C -
H bonds. $\Delta H ^{\circ}$ combustion (ethane) $=-372.0$
$\Delta H ^{\circ}$ combustion (propane) $=-530.0 \Delta H ^{\circ}$ for $C ($ graphite $) \rightarrow C ( g )=172.0$ Bond energy of $H - H =104.0 \Delta_{ f } H ^{\circ}$ of $H _2 O ( l )=-68.0 \Delta H ^{\circ}$ for $CO _2(g)=-94.0$
AnswerWe are given,
- $\text{C}_2\text{H}_6(\text{g})+\frac{7}{2}\text{O}_2(\text{g})\overrightarrow{\ \ \ \ \ }\ 2\text{CO}_2(\text{g})+\text{H}_2\text{O};$ $\Delta\text{H}^\circ=-372.0\text{kcal}$
- $\text{C}_3\text{H}_8(\text{g})+5\text{O}_2(\text{g})\overrightarrow{\ \ \ \ \ \ }\ 3\text{CO}_2(\text{g})+4\text{H}_2\text{O};$ $\Delta\text{H}^\circ=-530.0\text{kcal}$
- $\text{C(s)}\overrightarrow{\ \ \ \ \ }\ \text{C(g)};$ $\Delta\text{H}^\circ=172.0\text{kcal}$
- $\text{H}_2(\text{g})\overrightarrow{\ \ \ \ \ }\ 2\text{H(g)};$ $\Delta\text{H}^\circ=104.0\text{kcal}$
- $\text{H}_2(\text{g})+\frac{1}{2}\text{O}_2(\text{g})\overrightarrow{\ \ \ \ \ \ }\ \text{H}_2\text{O(l)};$ $\Delta\text{H}^\circ=-68.0\text{kcal}$
- $\text{C(g)}+\text{C}_2(\text{g})\overrightarrow{\ \ \ \ \ }\ \text{CO}_2(\text{g});$ $\Delta\text{H}^\circ=-94.0\text{kcal}$
Suppose the bond energy of C-C bond = xkcal $mol^{-1}.$ and that of C-H bond = ykcal $mol^{-1}.$ Then for $C_2H_6 (g),$
$\ \ \ \ \ \ \ \ \text{H} \ \ \ \ \ \text{H}\\ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ |\\\text{H}-\text{C}-\text{C}-\text{H}\overrightarrow{\ \ \ \ \ }\ 2\text{C(g)}+6\text{H};\\ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \text{H}\ \ \ \ \text{H}$ $\Delta\text{H}=\text{x}+6\text{y}\dots(\text{vii})$
and for $C_3H_8(g)$; i.e.,
$\ \ \ \ \ \ \ \ \text{H} \ \ \ \ \ \text{H} \ \ \ \ \ \text{H}\\ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\ \ \ \ \ \ |\\\text{H}-\text{C}-\text{C}-\text{C}-\text{H}\overrightarrow{\ \ \ \ \ }\ 3\text{C(g)}+8\text{H(g)};\\ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \text{H} \ \ \ \ \ \text{H}\ \ \ \ \text{H}$ $\Delta\text{H}=2\text{x}+8\text{y}\dots(\text{viii})$
To get eq. (vii), operate eq. (i) + 2 × eq. (iii) + 3 × eq. (iv) - 3 × eq. (v) - 2 × eq. (vi).
It gives $\Delta\text{H}=676\text{kcal}$
It get eq. (viii) operate eq. (ii) + eq. (iii) + 4 × eq. (iv) - 4 × eq. (v) - 3 × eq. (vi)
It gives $\Delta\text{H}=956\text{kcal}$
Thus, x + 6y = 676, 2x + 8y = 956
On solving these equations, we get x = 82, y = 99
Hence, C-C bond energy = 82kcal $mol^{-1}$ and C-H bond energy = 99kcal $mol^{-1}.$ View full question & answer→Question 453 Marks
Standard molar enthalpy of formation, $\Delta_\text{f}\text{H}^\ominus$ is just a special case of enthalpy of reaction, $\Delta_\text{r}\text{H}^\ominus$. Is the $\Delta_\text{r}\text{H}^\ominus$ for the following reaction same as $\Delta_\text{f}\text{H}^\ominus$? Give reason for your answer.$\text{CaO}(\text{s})+\text{CO}_2(\text{g})\rightarrow\text{CaCO}_3(\text{s});$ $\Delta_\text{f}\text{H}^\ominus=-178.3\text{kJ}\ \text{mol}^{-1}$
The value of $\Delta_\text{f}\text{H}^\ominus$ for $NH_3$ is $-91.8kJ mol^{-1}$. Calculate enthalpy change for the following reaction:$2\text{NH}_3(\text{g})\rightarrow\text{N}_2(\text{g})+3\text{H}_2(\text{g})$
AnswerNo, standard molar enthalpy of formation is the enthalpy of the reaction when 1mole of compound from its consituent elements.$\text{Ca}(\text{s})+\text{C}(\text{s})+\frac{3}{2}\text{O}_2(\text{g})\rightarrow\text{CaCO}_3(\text{s});$ $\Delta_\text{f}\text{H}^\circ$
AS the close reaction is different from the given reaction,$\therefore\ \Delta_\text{r}\text{H}^\circ\neq\Delta_\text{f}\text{H}^\circ$
View full question & answer→Question 463 Marks
The enthalpy of vaporisation of liquid diethyl ether $\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{O}$ is $26.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at its boiling point (35.0°C). Calculate $\Delta\text{S}^\circ$ for the conversion of:
- Liquid to vapour.
- Vapour to liquid at 35°C.
Answer
- For vaporisation of diethyl ether,
$\therefore\Delta_\text{vap}\text{S}^\circ=\frac{\Delta_\text{vap}\text{H}^\circ}{\text{T}}$
$\Delta_\text{vap}\text{H}^\circ=26.0\text{kJ mol}^{-1},$
$\text{T}=273+35=308\text{K}$
$\Delta_\text{vap}\text{S}^\circ=\frac{26.0\times10^3\text{J mol}^{-1}}{308\text{K}}$
$=84.4\text{JK}^{-1}\text{mol}^{-1}$
- The conversion of vapour into liquid is condensation. The enthalpy of condensation is negative of enthalpy of vaporisation.
$\therefore$ For condensation of diethyl ether (i.e., conversion of vapour to liquid)
$\Delta_\text{cond}\text{S}^\circ=\frac{\Delta_\text{cond}\text{H}^\circ}{\text{T}}$
$=\frac{-26.0\times10^3\text{mol}^{-1}}{308}$
$=-84.4\text{JK}^{-1}\text{mol}^{-1}$ View full question & answer→Question 473 Marks
It has been found that 221.4J is needed to heat 30g of ethanol from 15° to 18°. Calculate:
- specific heat capacity.
- molar heat capacity of ethanol.
Answer
- $\text{q} = \text{m} \times \text{specific heat} \times (\text{T}_{2} - \text{T}_{1})$
$\text{T}_{1} = 15 ^\circ\text{C} + 273 = 288 \text{K}$
$\text{T}_{2} = 18 ^\circ\text{C} + 273 = 291 \text{K}$
$221.4\text{J} = 30 \text{g} \times \text{specific heat} \times (291 \text{K} - 288 \text{K})$
$\text{Specific heat} = \frac{221.4}{30 \times 3} = 2.46\text{J} \text{ g}^{-1} \text{K}^{-1}$
- Molar heat capacity $=$ specific heat $\times$ Molar mass of $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
$= 2.46\text{J} \text{ g}^{-1} \text{K}^{-1} \times 46 \text{g} \text{ mol}^{-1}$
$= 113.16 \text{J K}^{-1} \text{mol}^{-1}$ View full question & answer→Question 483 Marks
An ideal gas is allowed to expand against a constant pressure of 2 bar from 10L to 50L in one step. Calculate the amount of work done by the gas. If the same expansion were carried out reversibly, will the work done be higher or lower than the earlier case? (Given that 1L bar = 100J)
AnswerWork done in the process can be calculated as$\text{w}=-\text{P}_\text{ex}\big(\text{V}_\text{f}-\text{V}_\text{t}\big)=-2\times40=-80\text{bar}=-8\text{kJ}$
The negative sign shows that work is done by the system on the surroundings. Work done will be more in the reversible expansion because internal pressure and external pressure are almost same at every step.
View full question & answer→Question 493 Marks
- Using the data given below, calculate the value of equilibrium constant for the reaction at 298K.
$3\text{CH}\equiv\text{CH}(\text{g})\rightleftharpoons\text{C}_6\text{H}_6(\text{g})$ assuming ideal gas behaviour, $\Delta_\text{f}\text{G}^\circ[\text{HC}\equiv\text{CH(g)}]=2.09\times10^5\text{J mol}^{-1},$ $\Delta_\text{f}\text{G}^\circ[\text{C}_6\text{H}_6(\text{g})]=1.24\times10^5\text{J mol}^{-1}, \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$.
- Based on your calculated value, comment whether this process can be recommended as a practical method for making benzene.
Answer
- $\Delta\text{G}^\circ=\Delta_\text{f}\text{G}^\circ[\text{C}_6\text{H}_6(\text{g})]-3\Delta_\text{f}\text{G}^\circ[\text{CH}\equiv\text{CH(g)}]$
$=1.24\times10^5\text{J mol}^{-1}-3\times2.09\times10^5\text{J mol}^{-1}$
$=1.24\times10^5\text{J mol}^{-1}-6.27\times10^5\text{J mol}^{-1}$
$=-5.03\times10^5\text{J mol}^{-1}$
$-5.03\times10^5=-2.303\times8.314\times298\log\text{K}$
$\Rightarrow\log\text{K}=\frac{5.03\times10^5\text{J}}{5705.8\text{J}}=+88.1559$
$\Rightarrow\text{K}=1.432\times10^{88}$
- Yes, this process can be recommended as a practical method for making benzene as value of K is very high.
View full question & answer→Question 503 Marks
In a process, 701J of heat is absorbed by a system and 394J of work is done by the system. What is the change in internal energy for the process?
AnswerAccording to the first law of thermodynamics,$\Delta\text{U} = \text{q} + \text{W} (\text{i})$
Where,$\Delta\text{U} =$ change in internal energy for a process
q = heat W = work Given, q = +701J (Since heat is absorbed) W = –394J (Since work is done by the system)Substituting the values in expression (i), we get
$\Delta\text{U} = 701\text{J} + (–394\text{J}) $
$\Delta\text{U} = 307\text{J}$
Hence, the change in internal energy for the given process is 307J.
View full question & answer→Question 513 Marks
Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from $35^{\circ} \mathrm{C}$ to $55^{\circ} \mathrm{C}$. Molar heat capacity of Al is $24 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$.
AnswerMass of Al = 60g Rise in temperature,$ \Delta\text{T} = 55 – 35 = 20°\text{C}$ Molar heat capacity of Al $= 24\text{J mol}^{-1} \text{K}^{-1}$ Specific heat capacity of Al $=\frac{24}{27}\text{Jg}^{-1} \ \text{K}^{-1}$$\therefore$ Energy required $ \text{m} \times \text{c} \times \Delta{\text{T}}$
= $60\times\frac{24}{27}\times20=\frac{28800}{27}=1066.67 \text{J}$
$= 1.068\text{kJ or 1.07kJ}$
View full question & answer→Question 523 Marks
The combustion of one mole of benzene takes place at 298K and 1atm. After combustion, $\mathrm{CO}_2(\mathrm{~g})$ and $\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ are produced and 3267.0kJ of heat is liberated. Calculate the standard enthalpy of formation of benzene $(\Delta_\text{f}\text{H}^\circ).$$\Delta_\text{f}\text{H}^\circ\text{CO}_2(\text{g})=-393\text{kJ mol}^{-1}$ and $\Delta_\text{f}\text{H}^\circ\text{H}_2\text{O(l})=-285.83\text{kJ mol}^{-1}.$
AnswerCombustion of 1 mole of benzene takes place as follows:$\text{C}_6\text{H}_6(\text{l})+\frac{15}{2}\text{O}_2\text{(g)}\overrightarrow{\ \ \ \ \ \ \ }\ 6\text{CO}_2\text{(g)}+3\text{H}_2\text{O(l)}$
$\Delta_\text{c}\text{H}^\circ=-3267.0\text{kJ mol}^{-1}$
$\Delta_\text{r}\text{H}^\circ=\Delta_\text{r}\text{H}^\circ$
$=\sum[\Delta_\text{f}\text{H}^\circ(\text{products})]-\sum[\Delta_\text{f}\text{H}^\circ(\text{reactants})]$
$\text{or} -3267.0\text{kJ mol}^{-1}$
$ -3267.0\text{kJ mol}^{-1}=6[\Delta_\text{f}\text{H}^\circ\text{CO}_2(\text{g})]\\+3[\Delta_\text{f}\text{H}^\circ\text{H}_2\text{O(l)}]-1[\Delta_\text{f}\text{H}^\circ\text{C}_6\text{H}_6(\text{l})]\\-\frac{15}{2}[\Delta_\text{f}\text{H}^\circ\text{C}_2(\text{g})]$
$ -3267.0\text{kJ}=(6)(-393.5\text{kJ mol}^{-1})\\+(3)(-285.83\text{kJ mol}^{-1})-(1)[\Delta_\text{f}\text{H}^\circ\text{C}_6\text{H}_6(\text{l})]-0$
$\Delta_\text{f}\text{H}^\circ\text{C}_6\text{H}_6(\text{l})=[6(-393.5)\\+3(-285.83)+3267]\text{kJ mol}^{-1}$
$=[-2361.0-857.49+3267.0]\text{kJ mol}^{-1}$
$=48.51\text{kJ mol}^{-1}$
View full question & answer→Question 533 Marks
Two moles of an ideal gas initially at 27°C and one atmospheric pressure are compressed isothermally and reversibly till the final pressure of the gas is 10atm. Calculate q, W and $\Delta\text{U}$ for the process.
AnswerHere, $\mathrm{n}=2$ moles, $\mathrm{T}=27^{\circ} \mathrm{C}=300 \mathrm{~K}, \mathrm{P}_1=1 \mathrm{~atm}, \mathrm{P}_2=10 \mathrm{~atm}$
$\text{W}=2.303\text{nRT}\log\frac{\text{p}_2}{\text{p}_1}$
$=2.303\times2\times8.314\text{JK}^{-1}\text{mol}^{-1}\times300\text{K}\\\times\log\frac{10}{1}=11488.28\text{J}$
For isothermal compression of ideal gas, $\Delta\text{U}=0$ Further, $\Delta\text{U}=\text{q}+\text{W}$$\therefore\text{q}=-\text{W}=-11488\text{J}$
View full question & answer→Question 543 Marks
The heat of combustion of $\mathrm{C}_2 \mathrm{H}_6$ is -368.4 kcal . Calculate heat of combustion of $\mathrm{C}_2 \mathrm{H}_4$, heat of combustion of $\mathrm{H}_2$ is $68.32 \mathrm{kcal} \mathrm{mol}{ }^{-1} . \Delta \mathrm{H}$ for the following reaction is $-37.1 \mathrm{kcal} . \mathrm{C}_2 \mathrm{H}_4(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \longrightarrow \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})$
Answer$\text{C}_2\text{H}_4(\text{g})+\text{H}_2(\text{g})\overrightarrow{\ \ \ \ \ \ }\ \text{C}_2\text{H}_6(\text{g});$$\Delta\text{H}=-37.1\text{kcal}$
$\Delta_\text{c}\text{H}^{\ominus}\text{C}_2\text{H}_6=-368.4\text{kcal},$ $\Delta_\text{c}\text{H}^\ominus\text{C}_2\text{H}_4=?$
$\Delta_\text{c}\text{H}^\ominus\text{H}_2(\text{g})=-68.32\text{kcal}$
$\Delta\text{H}=\sum\Delta_\text{c}\text{H}^\ominus(\text{reactants})-\sum\Delta_\text{c}\text{H}^\ominus(\text{products})$
$\Delta\text{H}=\Delta_\text{c}\text{H}^\ominus\text{C}_2\text{H}_4+\Delta_\text{c}\text{H}^\ominus\text{H}_2(\text{g})\\-\Delta_\text{c}\text{H}^\ominus\text{C}_2\text{H}_6(\text{g})$
$-37.1\text{kcal}=\Delta_\text{c}\text{H}^\ominus\text{C}_2\text{H}_4-68.32-(-368.4)$
$\Delta_\text{c}\text{H}^\ominus\text{C}_2\text{H}_4=-337.18\text{kcal}$
View full question & answer→