Question 12 Marks
The equilibrium constant for a reaction is 10. What will be the value of $\Delta\text{G}^\ominus$? $\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \mathrm{~T}=300 \mathrm{~K}$.
Answer$\Delta\text{G}^\ominus=-\text{RT ln k}=-2.303\text{RT}\log\text{K}.$$\text{R}=8.314\text{Jk}^{-1} \ \text{mol}^{-1};\text{T}=300\text{K};\text{K}=10$
$\Delta\text{G}^\ominus=-2.303\times8.314\text{Jk}^{-1} \ \text{mol}^{-1}\times(300\text{k})\times\log10 $
$=-5527\text{J} \ \text{mol}^{-1}=-5.527\text{kJ} \ \text{mol}^{-1}.$
View full question & answer→Question 22 Marks
The enthalpy of atomisation of $\mathrm{CH}_4$ is $1665 \mathrm{~kJ} \mathrm{~mol}^{-1}$. What is bond enthalpy of $\mathrm{C}-\mathrm{H}$ bond.
AnswerBond enthalpy of C-H bond $=\frac{1665}{4}=416.25\text{ kJ mol}^{-1}$$\because$ $\mathrm{CH}_4$ has 4 C-H bonds.
View full question & answer→Question 32 Marks
Calculate the standard free energy change $\Delta \mathrm{G}^{\circ}$ for the reaction: $2 \mathrm{HgO}(\mathrm{s}) \rightarrow 2 \mathrm{Hg}(\mathrm{l})+\mathrm{O}_2(\mathrm{~g}) \Delta \mathrm{H}^{\circ}=91 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $298 \mathrm{~K}^2 \mathrm{~S}^{\circ}(\mathrm{HgO})=72.0 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \mathrm{~S}_{(\mathrm{Hg})}^{\circ}=77.4 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ and $\mathrm{S}_{\left(\mathrm{O}_2\right)}^{\circ}=205 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ at 298 K .
Answer$\Delta\text{S}=2\text{S}^\circ_{(\text{Hg})}+\text{S}^\circ_{(\text{O}_2)}-2\text{S}^\circ_{(\text{HgO})}$$=(2\times77.4+205-2\times72.0)\text{J K}^{-1}\text{mol}^{-1}$
$=(154.8+205-144.0)\text{J K}^{-1}\text{mol}^{-1}$
$=(359.8-144.0)=215.8\text{J K}^{-1}\text{mol}^{-1}$
$\Delta_\text{r}\text{G}^\circ=\Delta_\text{r}\text{H}^\circ-\text{T}\Delta_\text{r}\text{S}^\circ$
$=91\text{kJ mol}^{-1}-\frac{298\times215.8}{1000}\text{kJ mol}^{-1}$
$=(91-64.3)\text{kJ mol}^{-1}=26.69\text{kJ mol}^{-1}$
View full question & answer→Question 42 Marks
The standard heat of formation of $\mathrm{CH}_4(\mathrm{~g}), \mathrm{CO}_2(\mathrm{~g})$ and $\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ are $-76.2,-394.8$ and -241.6 kJ mol ${ }^{-1}$ respectively. Calculate the amount of heat evolved by burning $1 \mathrm{~m}^3$ of methane measured at NTP.
AnswerThe burning of methane may be expressed as,$\text{CH}_4(\text{g})+2\text{O}_2(\text{g})\overrightarrow{\ \ \ \ \ \ } \ \text{CO}_2(\text{g})+2\text{H}_2\text{O(g)}$
$\Delta_\text{f}\text{H}^\circ=[\Delta_\text{f}\text{H}^\circ(\text{CO}_2)+2\Delta_\text{f}\text{H}^\circ(\text{H}_2\text{O})]\\-[\Delta_\text{f}\text{H}^\circ(\text{CH}_4)+2\Delta_\text{f}\text{H}^\circ(\text{O}_2)]$
$=[-394.8+2\times(-241.6)]-[-76.2+2\times0]$
$=-801.8\text{kJ}$
1 mole or 22.4 L of $\mathrm{CH}_4$ evolve heat $=801.8 \mathrm{~kJ} 1 \mathrm{~m}^3$ or 1000 L of $\mathrm{CH}_4$ evolve heat $=\frac{801.8 \times 1000}{22.4}=35794.6 \mathrm{~kJ}$
View full question & answer→Question 52 Marks
Standard vaporisation enthalpy of benzene at boiling point is $30.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$. For how long would 100 W electric heater have to operate in order to vaporise a 100 g sample at that temperature (power = energy/ time and $1 \mathrm{~W}=1 \mathrm{~J} \mathrm{~s}^{-1}$ )?
Answer$\Delta_{\text {vap }} \mathrm{H}^{\circ}$ (benzene $)=30.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ Molar mass of benzene, $\mathrm{C}_6 \mathrm{H}_6=(6 \times 12+6 \times 1) \mathrm{g} \mathrm{mol}^{-1}=78 \mathrm{~g} \mathrm{~mol}^{-1}$ Energy needed to vaporise benzene $=30.8 \mathrm{~kJ} \mathrm{~mol}^{-1} \times \frac{100 \mathrm{~g}}{78 \mathrm{~g} \mathrm{~mol}^{-1}}$
$=39.49 \mathrm{~kJ}$
So, Time $=\frac{\text { energy }}{\text { power }}=\frac{39.49 \mathrm{~kJ}, \mathrm{~J}}{100 \mathrm{~W}}$
$=\frac{39.49 \times 10^3 \mathrm{~J}}{100 \mathrm{~J} \mathrm{~s}^{-1}}=394.9 \mathrm{~s}=6.6 \mathrm{~min}$
View full question & answer→Question 62 Marks
Choose the correct answer. For the process to occur under adiabatic conditions, the correct condition is:
- $\Delta\text{T}=0$
- $\Delta\text{P}=0$
- $\text{q}=0$
- $\text{w}=0$
Answer
- $\Delta\text{P}=0$
Explanation:
A system is said to be under adiabatic conditions if there is no exchange of heat between the system and its surroundings. View full question & answer→Question 72 Marks
Predict in which of the following, entropy increases/ decreases. Give reason.
- Temperature of crystalline solid is raised from 0K to 115K.
- $\mathrm{H}_2(\mathrm{g}) \rightarrow 2 \mathrm{H}(\mathrm{g})$
Answer
- Entropy will increase on increasing the temperature since the particles of solid move with greater speed at higher temperature. At 0K, there is perfect order of the constituent particles, entropy is minimum, tends to zero.
- $\mathrm{H}_2(\mathrm{g}) \rightarrow 2 \mathrm{H}(\mathrm{g})$
Entropy will increase because the number of particles of product are double than that of reactant. View full question & answer→Question 82 Marks
What are the correct thermodynamic conditions for the spontaneous reaction at all temperature?
Answer$\Delta\text{H}^\circ<0,\Delta\text{S}^\circ<0.$$\because\Delta\text{G}^\circ=\Delta\text{H}^\circ-\text{T}\Delta\text{S}^\circ,$
$\text{If }\Delta\text{H}^\circ=-\text{ve},\Delta\text{S}=+\text{ve},\Delta\text{G}^\circ$ will be negative for all values of 'T'.
View full question & answer→Question 92 Marks
Define the term entropy. Write its unit. How does entropy of a system change on increasing temperature?
AnswerEntropy is measure of degree of disorder or randomness. Its unit is $\mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. Entropy of system increases with increase in temperature.
View full question & answer→Question 102 Marks
For oxidation of iron,$4\text{Fe(s)}+3\text{O}_2(\text{g})\overrightarrow{\ \ \ \ \ \ }\ 2\text{Fe}_2\text{O}_3(\text{s})$
entropy change is $-549.4 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ at 298 K . Inspite of negative entropy change of this reaction, why is the reaction spontaneous? $\left(\Delta_{\mathrm{r}} \mathrm{H}^{\circ}\right.$ for this reaction is $\left.-1648 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}\right)$
Answer$4\text{Fe(s)}+3\text{O}_2(\text{g})\overrightarrow{\ \ \ \ \ \ }\ 2\text{Fe}_2\text{O}_3(\text{s})$$\Delta\text{S}_\text{surr}=-\frac{\Delta_\text{r}\text{H}^\circ}{\text{T}}=\frac{-(-1648\times10^3\text{ J mol}^{-1})}{289\text{K}}$
$\Delta\text{S}_\text{surr}=5530\text{J K}^{-1}\text{mol}^{-1}$
$\Delta\text{S}_\text{sys}=-549.4\text{J K}^{-1}\text{mol}^{-1}$
$\Delta_\text{r}\text{S}_\text{total}=\Delta\text{S}_\text{sys}+\Delta\text{S}_\text{sys}$
$\Rightarrow\Delta_\text{r}\text{S}_\text{total}=5530-549.4$
$=4980.6\text{J K}^{-1}\text{mol}^{-1}$
Since $\Delta_\text{r}\text{S}_\text{total}$ is +ve, therefore, reaction is spontaneous.
$\because\Delta\text{G}=-\text{T}\Delta\text{S}_\text{total}$
$\because\Delta\text{G}=-\text{ve}$
When $\text{S}_\text{total}$ is positive.
View full question & answer→Question 112 Marks
Calculate the free energy change when 1 mole of NaCl is dissolved in water at 298 K . (Given, lattice energy of $\mathrm{NaCl}=$ $-777.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$, hydration energy $=-774.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta \mathrm{S}=0.043 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ at 298 K ).
Answer$\Delta\text{H}=$ hydration energy - lattice energy$\Delta\text{H}=-774.1\text{kJ mol}^{-1}-(-777.8\text{kJ mol}^{-1})\\=3.7\text{kJ mol}^{-1}$
$\Delta\text{G}=\Delta\text{H}-\text{T}\Delta\text{S}$
$=+3.7-298\times0.043=+3.7-12.81$
$\Delta\text{G}=-9.11\text{kJ mol}^{-1}$
View full question & answer→Question 122 Marks
Give one point to differentiate the following thermodynamic terms:
- Extensive properties and intensive properties.
- Isothermal process and isobaric process.
Answer
-
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Extensive Property
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Intensive Property
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The property that depends on the quantity of a matter contained in the system, e.g. mass, volume and heat capacity.
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These are the properties that depends on the nature of the substance and not on the amount of substance, e.g. refractive index and viscosity.
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-
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Isothermal Process
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Isobaric Process
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When a process is carried out in such a manner that the temperature remains constant, it is called isothermal process.
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Isobaric process is the one during which the pressure of the system remains constant.
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View full question & answer→Question 132 Marks
What will be work done during expansion of a gas from $4 \mathrm{dm}^3$ to $6 \mathrm{dm}^3$ against an external pressure of 3atm. (1L atm $=101.32 \mathrm{~J})$
Answer$P_{\text {ext }}=3 \mathrm{~atm}, V_f=6 \mathrm{dm}^3, V_i=3 \mathrm{~d} \mathrm{~cm}^3 \mathrm{w}=-P \Delta V$
$=-3 \times(6-4)=-6 \mathrm{L~atm}=-6 \times 101.32 \mathrm{~J}=-607.92 \mathrm{~J}=-608 \mathrm{~J}$
View full question & answer→Question 142 Marks
The enthalpy of reaction for the reaction : $2\text{H}_2(\text{g})+\text{O}_2(\text{g})\rightarrow2\text{H}_2\text{O}(\text{l})\ \text{is}\ \Delta_\text{r}\text{H}^\ominus=-572\text{kJ}\ \text{mol}^{-1}$
What will be standard enthalpy of formation of $H_2O(l)$?
AnswerAccording to the definition of standard enthalpy of formation, the enthalpy change for the following reaction will be standard enthalpy of formation of $H_2O (l)\text{H}_2(\text{g})+\frac{1}{2}\text{O}_2(\text{g})\rightarrow\text{H}_2\text{O}(\text{l}).$
or the standard enthalpy of formation of $H_2O(l)$ will be half of the enthalpy of the given equation i.e., $\Delta_\text{r}\text{H}^\ominus$ is also halved.$\Delta_\text{f}\text{H}^\ominus_{\text{H}_2\text{O}}(\text{l})=\frac{1}{2}\times\Delta_\text{r}\text{H}^\ominus=\frac{-572\text{kJ}\ \text{mol}^{-1}}{2}=-286\text{kJ}\ \text{mol}^{-1}$
View full question & answer→Question 152 Marks
Enthalpy is an extensive property. In general, if enthalpy of an overall reaction A → B along one route is $\Delta_\text{r}\text{H}$ and $\Delta_\text{r}\text{H}_1,\Delta_\text{r}\text{H}_2,\Delta_\text{r}\text{H}_3......$ represent enthalpies of intermediate reactions leading to product B. What will be the relation between $\Delta_\text{r}\text{H}$ for overall reaction and $\Delta_\text{r}\text{H}_1,\Delta_\text{r}\text{H}_2,.....$etc. for intermediate react.
Answer
$\Delta_\text{r}\text{H}=\Delta_\text{r}\text{H}_1+\Delta_\text{r}\text{H}=\Delta_\text{r}\text{H}_2+\Delta_\text{r}\text{H}=\Delta_\text{r}\text{H}_3....$
It can be represented as:
View full question & answer→Question 162 Marks
18.0 g of water completely vapourises at $100^{\circ} \mathrm{C}$ and 1 bar pressure and the enthalpy change in the process is 40.79 kJ $\mathrm{mol}^{-1}$. What will be the enthalpy change for vapourising two moles of water under the same cond tions? What is the standard enthalphy of vapourisation for water?
AnswerEnthalpy of a reaction is the energy change per mole for the process.
$18 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}=1 \mathrm{~mole}\left(\Delta \mathrm{H}_{\text {vap }}=40.79 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$
Enthalpy change for vapourising 2 moles of $\mathrm{H}_2 \mathrm{O}=2 \times 40.79=81.58 \mathrm{~kJ} \Delta \mathrm{H}_{\text {vap }}^{\circ}=40.79 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}$
View full question & answer→Question 172 Marks
Heat capacity $\left(C_p\right)$ is an extensive property but specific heat (c) is an intensive property. What will be the relation between $C_p$ and $c$ for 1 mol of water?
AnswerFor water, molar heat capacity $=18 \times$ Specific heat or
$C_p=18 \times c$
But, specific heat,
$C=4.18 \mathrm{Jg}^{-1} \mathrm{~K}^{-1} \text { Heat capacity, }$
$C_p=18 \times 4.18 \mathrm{JK}^{-1}=75.24 \mathrm{JK}^{-1}$
View full question & answer→Question 182 Marks
Establish a relationship between $\Delta\text{H}$ and $\Delta\text{U}$ in Haber's process of synthesis of ammonia assuming that gaseous reactants and products are ideal.
AnswerHaber's process of synthesis for ammonia is,$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\overrightarrow{\ \ \ \ \ \ }\ 2\text{NH}_3(\text{g})$
$\Delta\text{n}_\text{g}=2-(1+3)=-2$
But $\Delta\text{H}=\Delta\text{U}+\Delta\text{n}_\text{g}\text{RT}$$\therefore\Delta\text{H}=\Delta\text{U}-2\text{RT}$
View full question & answer→Question 192 Marks
Consider the following reaction: $\mathrm{H}_2(\mathrm{~g})+\mathrm{F}_2(\mathrm{~g}) \rightarrow 2 \mathrm{HF}(\mathrm{g}) \Delta\text{H}^\circ=-542\text{kJ},\ \Delta\text{S}^\circ=14\text{JK}^{-1}$
Calculate the $\Delta\text{G}^\circ$ value for the reaction and state if the reaction is spontaneous at 298K.
Answer$\Delta\text{G}^\circ=\Delta\text{H}^\circ-\text{T}\Delta\text{S}^\circ$$\Delta\text{G}^\circ=-542\text{kJ}-\frac{298\times14}{1000}\text{kJ}$
$=-542-4=-546\text{kJ}$
Since $\Delta\text{G}^\circ$ is negative, therefore, reaction is spontaneous at 298K.
View full question & answer→Question 202 Marks
The difference between $C_p$ and $C_V$ can be derived using the empirical relation $H=U+p V$ Calculate the difference between $C_p$ and $C_V$ for 10 moles of an ideal gas.
AnswerGiven that, $C_v=$ heat capacity at constant volume, $C_p=$ heat capacity at constant pressure Difference between $C_p$ and $C_v$ is equal to gas constant $(R) . \therefore C_p-C_v=n R$ (where, $n=$ no. of moles)
$=10 \times 8.314=83.14 \mathrm{~J}$
View full question & answer→Question 212 Marks
Why standard entropy of an elementary substance is not zero whereas standard enthalpy of formation is taken as zero?
AnswerA substance has a perfectly ordered arrangement of its constituent particles only at absolute zero. Hence, entropy is zero only at absolute zero. Enthalpy of formation is the heat change involed in the formation of one mole of the substance from its elements. An element formed from itself means no heat change, i.e. $\Delta_\text{f}\text{H}^\circ=0.$
View full question & answer→Question 222 Marks
Increase in enthalpy of the surroundings is equal to decrease in enthalpy of the system. Will the temperature of system and surroundings be the same when they are in thermal equilibrium?
AnswerYes, If the system is in thermal equillibrium with the surroundings, then the tempreature of the surroundings is same as that of the system. Also, increase in enthalpy the surroundigs is equal to decrease in the enthalpy of the system.
View full question & answer→Question 232 Marks
Which has more entropy? 1 mole $\mathrm{H}_2 \mathrm{O}(\mathrm{I})$ at $25^{\circ} \mathrm{C}$ or 1 mole $\mathrm{H}_2 \mathrm{O}(\mathrm{I})$ at $35^{\circ} \mathrm{C}$?
Answer1 mole of $\mathrm{H}_2 \mathrm{O}$ at $35^{\circ} \mathrm{C}$ has more entropy because disorder or randomness increases with increase in temperature.
View full question & answer→Question 242 Marks
A 1.25 g sample of octane $\left(\mathrm{C}_8 \mathrm{H}_{18}\right)$ is burnt in excess of oxygen in a bomb calorimeter. The temperature of calorimeter rises from 294.05 to 300.78 K . If heat capacity of the calorimeter is 8.93 kJ K . Find heat transferred to calorimeter.
Answer$\text{m}=\text{n}\times\text{c}\times(\text{T}_2-\text{T}_1)$$=\frac{\text{mass}}{\text{molar mass}}\times\text{C}\times(\text{T}_2-\text{T}_1)$
$=\frac{1.25}{114}\times8.93\text{kJ K}^{-1}\times(300.78\text{K}-294.05\text{K})$
$=\frac{1.25\times8.93\times6.73}{114}=\frac{75.124}{124}$
$=0.66\text{kJ mol}^{-1}$
View full question & answer→Question 252 Marks
Given that $\Delta\text{H}=0$ for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?
AnswerYes it will be a spontaneous process since delta G will be negative and entropy will be positive after diffusing two gases (volume will increase).$\Delta\text{G}=\Delta\text{H}-\text{T}\Delta\text{S}$
View full question & answer→Question 262 Marks
The combustion of benzene(I) gives $\mathrm{CO}_2(\mathrm{g})$ and $\mathrm{H}_2 \mathrm{O}(\mathrm{I})$. Given that heat of combustion at constant volume is $-3263.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $25^{\circ} \mathrm{C}$; calculate heat of combustion in $\mathrm{kJ} \mathrm{~mol}{ }^{-1}$ at constant pressure. $\left[\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right]$
Answer$\text{C}_6\text{H}_6(\text{l})+\frac{15}{2}\text{O}_2(\text{g})\overrightarrow{\ \ \ \ \ }\ 6\text{CO}_2(\text{g})+3\text{H}_2\text{O(l)}$ $\Delta\text{H}=?$$\Delta\text{n}=\sum(\text{n}_\text{g})_\text{products}-\sum(\text{n}_\text{g})_\text{reactants}$
$\Delta\text{n}=6-\frac{15}{2}=-\frac{3}{2}$ $[\text{T}=25^\circ\text{C}+273=298\text{K}]$
$\Delta\text{H}=\Delta\text{U}+\Delta\text{nRT}$
$=-3263.9-\frac{3}{2}\times\frac{8.314\times298\text{K}}{1000}$
$=-3263.90\text{kJ}-\frac{3716.36}{1000}\text{kJ}$
$=-3263.92\text{kJ}-3.72\text{kJ}$
$=-3267.62\text{kJ mol}^{-1}$
View full question & answer→Question 272 Marks
Calculate the electron gain enthalpy of fluorine from the data given below: $\Delta_{\mathrm{f}} \mathrm{H}^{\circ}$ of $\mathrm{KF}=-560.8 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}$, Sublimation energy of $\mathrm{K}=87.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$, Dissociation energy of F , is $158.9 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}$, Lattice energy of $\mathrm{KF}^{-15}-807.5 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}$ and Ionisation energy of K is $414.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
AnswerElectron gain enthalpy $=\Delta_\text{f}\text{H}-\text{S}-\frac{1}{2}\text{D}-\text{I.E.}-\text{U}$ 'S' is sublimation energy; 'D' is bond dissociation enthalpy. I.E. is ionisation elthalpy. 'U' represents lattice energy.$\text{E.G.E.}=-560.8-87.8-\frac{1}{2}\\\times158.9-414.2-(-807.5)$
$\text{E.G.E.}=-1142.25+807.5\\=-334.75\text{kJ mol}^{-1}$
View full question & answer→Question 282 Marks
The value of $\Delta_\text{f}\text{H}^\ominus$ for $\mathrm{NH}_3$ is $-91.8 \mathrm{~kJ} \mathrm{~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})$
Answer$\frac{1}{2}\text{N}_{2_{(\text{g})}}+\frac{3}{2}\text{H}_{2_{(\text{g})}}\rightarrow\text{NH}_{3_{(\text{g})}},$ $\Delta_\text{f}\text{H}^\circ=-91.8\text{kJ}\ \text{mol}^{-1}$or $\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightarrow2\text{NH}_3(\text{g}),$
$\Delta_\text{f}\text{H}^\circ=-2\times91.8=-183.6\text{kJ}\ \text{mol}^{-1}$
SO, for the reverce reaction,
$\Delta_\text{r}\text{H}^\circ=183.6\text{kJ}\ \text{mol}^{-1}$
$\therefore\text{The}\ \text{value}\ \text{of}\ \Delta _\text{r}\text{H}^\circ\ \text{for}\ \text{NH}_3\ \text{is}\ +183.6\text{kJ}\ \text{mol}^{-1}$
View full question & answer→Question 292 Marks
Calculate the temperature above which the reduction of lead oxide in the following reaction becomes spontaneous:
PbO(s) + C(s) → Pb(s) + CO(g)
Given: $[\Delta\text{H}=108.4\text{kJ mol}^{-1};\ \Delta\text{S}=190\text{J K}^{-1}\text{mol}^{-1}]$
Answer$\Delta\text{G}=\Delta\text{H}-\text{T}\Delta\text{S}$At equilibrium, $\Delta\text{G}=0\Rightarrow\Delta\text{H}=\text{T}\Delta\text{S}$
$\Rightarrow\text{T}=\frac{\Delta\text{H}}{\Delta\text{S}}=\frac{108.4\times1000\text{ J mol}^{-1}}{190\text{JK}^{-1}\text{mol}^{-1}}$
$\text{T}=570.526\text{K}$
The reaction will become spontaneous above 570.52K because $\Delta\text{G}$ is equal to zero at this temperature and above this temperature, $\Delta\text{G}$ will become -ve.
View full question & answer→Question 302 Marks
Give reasons for the following:
- The enthalpy of neutralisation is always constant, i.e. 57.1kJ/ mol when a strong acid neutralism a strong base.
- Neither q nor w is a state function but q + w is a state function.
Answer
- It is because strong acid and strong base are completely ionised in aqueous solution.
- $\Delta\text{U}=\text{q}+\text{w}$
Since, $\Delta\text{U}$ is a state function and it is equal to q + w which is state function, because $\Delta\text{U}$ depends upon initial and final states and not on path. View full question & answer→Question 312 Marks
Find out whether it is possible to reduce MgO using carbon at 298 K. If not, at what temperature it becomes spontaneous? For reaction:$\text{MgO(s)}+\text{C(s)}\overrightarrow{\ \ \ \ \ \ }\ \text{Mg(s)}+\text{CO(g)}$
$\Delta_\text{r}\text{H}^\circ=91.18\text{kJ mol}^{-1}$ and $\Delta_\text{r}\text{S}^\circ=197.67\text{kJ mol}^{-1}$
Answer$\Delta_\text{r}\text{G}^\circ=\Delta_\text{r}\text{H}^\circ-\text{T}\Delta_\text{r}\text{S}^\circ$$=91.18\text{kJ mol}^{-1}-\frac{298\text{K}\times197.67}{1000}\text{kJ K}^{-1}\text{mol}^{-1}$
$=91.18\text{kJ mol}^{-1}-58.91\text{kJ mol}^{-1}$
$=32.27\text{kJ mol}^{-1}$
Since $\Delta\text{G}$ is +ve, therefore, carbon cannot reduce MgO.
$\Delta_\text{r}\text{G}^\circ=\Delta_\text{r}\text{H}^\circ-\text{T}\Delta_\text{r}\text{S}^\circ$
$0=91.18\text{kJ mol}^{-1}\\-\text{T}\times\frac{197.67}{1000}\text{kJ K}^{-1}\text{mol}^{-1}$
$\text{T}=\frac{91.18\times1000}{197.67}$
$=\frac{91180}{197.67}=461.27\text{K}$
Above 461.27K, $\Delta\text{G}=-\text{ve},$ reaction will become spontaneous.
View full question & answer→Question 322 Marks
Predict the sign of $\Delta\text{S}$ for the following:
- A sample of iron cools from 50°C to 25°C.
- HgO thermally decomposes to $\mathrm{Hg}(\mathrm{l})$ and $\mathrm{O}_2(\mathrm{g})$.
Answer
- $\Delta\text{S}=-\text{ve}.$
- $\Delta\text{S}=+\text{ve}.$
View full question & answer→Question 332 Marks
Consider the following gas phase reaction:$\text{H}_{2}\text{(g)} + \text{Cl}_{2} \text{(g)} \rightleftharpoons\text{2HCl(g)}$
$\Delta\text{H}^{\circ} = -92\text{kJ}, \ \Delta\text{S}^{\circ} = -95\text{JK}^{-1}$
Calculate the equilibrium constant, $K_p$ at 298 K for this reaction.
Answer$\Delta \text{G}^{\circ} =\Delta \text{H}^{\circ} - \text{T}\Delta\text{S}^{\circ}$$ = 92\text{kJ} \frac{298 \times 95}{1000}\text{kJ} = -64\text{kJ}$
$\Delta \text{G}^{\circ} = -2.303 \text{RT} \log \text{k}_{\text{p}}$
$-64\text{kJ} = \frac{-2.303 \times 8.314\times 298\text{K} }{1000}\log \text{K}_{\text{p}}$
$\log \text{K}_{\text{sp}} = \frac{-64\text{kJ}}{-5.7058\text{kJ}} = 11.2166$
$\text{K}_{\text{p}} = \text{Antilog } 11.2166$
View full question & answer→Question 342 Marks
Which has larger absolute entropy per mole?
i. $\mathrm{H}_2 \mathrm{O}(\mathrm{I})$ at 298 K or $\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ at 350 K
ii. $\mathrm{N}_2$ or NO both at 298 K
Answer
- $\mathrm{H}_2 \mathrm{O}$ at 350K has larger absolute entropy per mole as entropy (randomness) is more at higher temperature.
- NO at 298K has greater entropy because it is unstable, highly reactive.
View full question & answer→Question 352 Marks
- Name the two factors that influence the entropy change that takes place in the surrounding.
- Write an equation that shows the relationship.
Answer
- $\Delta\text{S}_\text{surr}$ is directly proportional to $-q_p$, and inversely proportional to temperature in Kelvin.
$-\text{q}_\text{p}=-\Delta\text{H}_\text{sys}$
- $\Delta\text{S}_\text{surr}=\frac{-\Delta\text{H}_\text{sys}}{\text{T}}$
View full question & answer→Question 362 Marks
Calculate $\Delta_\text{r}\text{H}^\circ$ for the reaction,
$\text{H}-\text{C}=\text{C}-\text{H}+30=0\overrightarrow{\ \ \ \ \ \ \ }\\ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \text{H} \ \ \ \ \ \text{H}\\20=\text{C}=0+2\text{H}-0-\text{H}$
The average bond enthalpies of various bonds are as follows.
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Bond
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Bond enthalpy ($\mathrm{kJ} \mathrm{~mol}^{-1}$)
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C-H
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414
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|
O=O
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499
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|
C=O
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724
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O-H
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460
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|
C=C
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619
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AnswerIn this reaction, four $\mathrm{C}-\mathrm{H}$ bonds, one $\mathrm{C}=\mathrm{C}$ bond and three $\mathrm{O}=\mathrm{O}$ bonds are broken and four $\mathrm{C}=\mathrm{O}$ bonds and four $\mathrm{O}-\mathrm{H}$ bonds are formed. Thus, $\Delta \mathrm{H}^{\circ}=[4 \times$ (bond enthalpy of $\mathrm{C}-\mathrm{H})+$ (bond enthalpy) of $\mathrm{C}=\mathrm{C}$ ) $+3[$ bond enthalpy of $\mathrm{O}=\mathrm{O}]$ $-4 \times[$ bond enthalpy of $\mathrm{C}=\mathrm{O}]+4$ (bond enthalpy of $\mathrm{O}-\mathrm{H}$ ) $]$
$\Delta \mathrm{H}=(4 \times 414+619+3 \times 499)-(4 \times 724+4 \times 460)$
$=-964 \mathrm{~kJ} \mathrm{~mol}^{-1}$
View full question & answer→Question 372 Marks
Heat has randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. Write the mathematical relation which relates these three parameters.
AnswerHeat has randomizing influence on a system and temperature is the measure of average chaotic motion of particles in the system. The mathematical relation which relates these three parameters is $\Delta\text{S}=\frac{\text{q}_\text{rev}}{\text{T}}$ Here, $\Delta\text{S}=$ change in entropy$\text{q}_\text{rev}=$ heat of reversible reaction
T = tempreature
View full question & answer→Question 382 Marks
Use the following thermodynamic data to calculate the enthalpy change for the formation of solid lithium fluoride, LiF(s) from Li(s) and $\text{F}_2$(g):$\text{Li(s)}+\frac{1}{2}\text{F}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ }\ \text{LiF(s)}$
$\text{Li(s)}\overrightarrow{\ \ \ \ \ \ \ \ }\ \text{Li(g)};$ $\Delta_\text{s}\text{H}^\circ=155\text{kJ mol}^{-1}$
$\frac{1}{2}\text{F}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ }\ \text{F(g)};$ $\Delta\text{H}^\circ=75\text{kJ}$
$\text{Li(g)}\overrightarrow{\ \ \ \ \ } \ \text{Li}^{+}(\text{g})+\text{e}^-;$ $\Delta\text{H}^\circ=520\text{kJ mol}^{-1}$
$\text{F(g)}+\text{e}^-\overrightarrow{\ \ \ \ \ \ } \ \text{F}^-(\text{g});$ $\Delta\text{H}^\circ=-333\text{kJ mol}^{-1}$
$\text{Li}^+\text{(g)}+\text{F}^-(\text{g})\overrightarrow{\ \ \ \ \ \ \ \ }\ \text{LiF(s)};$ $\Delta\text{H}^\circ=-1012\text{kJ mol}^{-1}$
Answer$\Delta_\text{f}\text{H}=\Delta_\text{s}\text{H}^\circ+\Delta_\text{D}\text{H}^\circ+\Delta_\text{I}\text{H}^\circ\\+\Delta_\text{EA}\text{H}^\circ+\Delta_\text{Lattice}\text{H}^\circ$$=155+75+520-333-1012$
$=750-1345=-595\text{kJ mol}^{-1}$
$\Delta_\text{s}\text{H}^\circ=$ enthalpy of sublimation
$\Delta_\text{D}\text{H}^\circ=$ Bond dissociation enthalpy
$\Delta_\text{I}\text{H}^\circ=$ Ionisation enthalpy
$\Delta_\text{EA}\text{H}^\circ=$ Electron gain enthalpy
$\Delta_\text{lattice}\text{H}^\circ=$ Lattice energy
$\Delta_\text{f}\text{H}^\circ=$ Enthalpy of formation
View full question & answer→Question 392 Marks
Comment on the spontaneity of a reaction at constant temperature and pressure in the following cases:
- $\Delta\text{H}<0\text{ and }\Delta\text{S}>0$
- $\Delta\text{H}>0\text{ and }\Delta\text{S}<0$
- $\Delta\text{H}<0\text{ and }\Delta\text{S}<0$
- $\Delta\text{H}>0\text{ and }\Delta\text{S}>0$
Answer
- $\Delta\text{H}<0\text{ and }\Delta\text{S}>0;$ $\Delta\text{G}$ will be -ve and process will always be spontaneous.
- $\Delta\text{H}>0\text{ and }\Delta\text{S}<0;$ $\Delta\text{G}$ will be +ve and process will never be spontaneous.
- $\Delta\text{H}<0\text{ and }\Delta\text{S}<0;$ $\Delta\text{G}$ will be -ve, if $\Delta\text{H}>\text{T}\Delta\text{S}$ and the process will be spontaneous at low temperature.
- $\Delta\text{H}>0\text{ and }\Delta\text{S}>0;$ $\Delta\text{G}$ will be -ve, if $\text{T}\Delta\text{S}>\Delta\text{H}$ and the process will be spontaneous at high temperature.
View full question & answer→Question 402 Marks
The enthalpy of vapourisation of $\mathrm{CCl}_4$ is $30.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Calculate the heat required for the vapourisation of 284 g of $\mathrm{CCl}_4$ at constant pressure. (Molar mass of $\mathrm{CCl}_4=154 \mathrm{~g} \mathrm{~mol}^{-1}$ ).
Answer$\text{q}_\text{p}=\Delta\text{H}=30.5\text{kJ}\ \text{mol}^{-1}$$\therefore$ Heat required for vapourisation of 284g of $\mathrm{CCl}_4$
$=\frac{284\text{g}}{154\text{g}\ \text{mol}^{-1}}\times30.5\text{kJ}\ \text{mol}^{-1}$
$= 56.2\text{kJ}$
View full question & answer→Question 412 Marks
A swimmer coming out from a pool is covered with a film of water weighing about 18 g . How much heat must be supplied to evaporate this water at 298 K ? Calculate the internal energy of vaporisation at $100^{\circ} \mathrm{C} . \Delta_{\text {vap }} \mathrm{H}^{\circ}$ for water at $373 \mathrm{~K}=40.66 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
Answer$18\text{g of H}_2\text{O(liquid)}\xrightarrow{\text{Vaporisation}}18\text{g H}_2\text{O(vapour)}$$\Delta_\text{vap}\text{H}^\circ\text{ for H}_2\text{O}=40.66\text{kJ mol}^{-1}$
No. of moles in 18g of water $=\frac{18\text{g}}{18\text{g mol}^{-1}}=1\text{ mole}$
1 mole of $\mathrm{H}_2 \mathrm{O}$ needs 40.66 kJ of energy for vaporization
$\Delta_\text{vap}\text{H}^\circ=\Delta_\text{vap}\text{U}^\circ+\Delta \text{nRT}$
$\Delta_\text{vap}\text{U}^\circ=\Delta_\text{vap}\text{H}^\circ-\Delta \text{nRT}$
$=40.66-1\times8.314\times10^{-3}\times373$
$=37.56\text{kJ mol}^{-1}$
View full question & answer→Question 422 Marks
At 0°C, ice and water are in equilibrium and $\Delta\text{H}=6.06\text{kJ mol}^{-1}$ for the process, $\mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})$
What will be $\Delta\text{S}$ and $\Delta\text{G}$ for the conversion of ice into liquid water?
AnswerFor equilibrium reaction,$\text{H}_2\text{O(s)}\rightleftharpoons\text{H}_2\text{O(l)}$
$\Delta\text{G}=0\Rightarrow\Delta\text{G}=\Delta\text{H}-\text{T}\Delta\text{S}$
$\Rightarrow0=\Delta\text{H}-\text{T}\Delta\text{S}$
$\Delta\text{S}=\frac{\Delta\text{H}}{\text{T}}=\frac{6.06\text{kJ mol}^{-1}}{273}$
$=0.02219\text{kJ K}^{-1}\text{mol}^{-1}$
$=22.2\text{JK}^{-1}\text{mol}^{-1}$
View full question & answer→Question 432 Marks
Acetic acid (ethanoic acid) and hydrochloric acid react with KOH solution. The enthalpy of neutralisation of ethanoic acid is $-55.8kJ mol^{-1}$ while that of hydrochloric acid is $-57.3kJ mol^{-1}$
Can you think of why are these different?
AnswerThe thermochemical equations for the neutralisation of acetic acid and hydrochloric acids are,$\text{CH}_3\text{COOH(aq)}+\text{KOH(aq)}\overrightarrow{\ \ \ \ \ \ }\ \text{CH}_3\text{COOK(aq)}\\+\text{H}_2\text{O(l)};$ $\Delta_\text{n}\text{H}^\circ=-55.8\text{kJ mol}^{-1}$
and $\text{HCl(aq)}+\text{KOH(aq)}\overrightarrow{\ \ \ \ \ }\ \text{KCl(aq)}+\text{H}_2\text{O(l)};$ $\Delta_\text{n}\text{H}^\circ=-57.3\text{kJ mol}^{-1}$ In dilute solution HCl is completely ionised into $H^+$ and $Cl^-$, whereas $CH_3COOH$ is ionised to very small extent. This difference between the enthalpy of neutralisation of the two acids is due to tie endothermic nature of the reaction involving tie ionisation of acetic acid, i.e.,$\text{CH}_3\text{COOH(aq)}\rightleftharpoons\text{CH}_3\text{COO}^-(\text{aq})+\text{H}^+(\text{aq})$
$\Delta_\text{ion}\text{H}^\circ=-55.8-(-57.3)\text{kJ mol}^{-1}$
$=1.5\text{kJ mol}^{-1}$
View full question & answer→Question 442 Marks
Calculate the standard enthalpy of formation of $\mathrm{C}_2 \mathrm{H}_4(\mathrm{~g})$ from the following thermochemical equation.
$\mathrm{C}_4 \mathrm{H}_8(\mathrm{~g})+6 \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_2+4 \mathrm{H}_2 \mathrm{O}$
Given that $\Delta_{\mathrm{f}} \mathrm{H}^{\ominus}$ of $\mathrm{CO}_2(\mathrm{~g}), \mathrm{H}_2 \mathrm{O}(\mathrm{g})$ as -393.5 and $-249 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively.
Answer$\Delta_\text{f}\text{H}^\ominus$ of $O_2(g) = 0$ by convention$\Delta_\text{r}\text{H}^-=\sum\Delta_\text{f}\text{H}^\ominus(\text{products})-\sum\Delta_\text{f}\text{H}^\ominus(\text{reactants})$
Substituting the given values
$-2646=[4(-393.5)+4\times(-249.0)]\\-[\Delta_\text{f}\text{H}^\ominus(\text{C}_4\text{H}_8)+0]$
$=[-1574-996]-\Delta_\text{f}\text{H}^\ominus(\text{C}_2\text{H}_4)$
$=-2570=\Delta_\text{f}\text{H}^\ominus(\text{C}_2\text{H}_4)$
$\Delta_\text{f}\text{H}^\ominus(\text{C}_2\text{H}_4)=2646-2570=76\text{kJ mol}^{-1}$
View full question & answer→Question 452 Marks
The equilibrium constant at $25^{\circ} \mathrm{C}$ for the process $\mathrm{CO}^{3+}(\mathrm{aq})+6 \mathrm{NH}_3(\mathrm{aq}) \rightleftharpoons\left[\mathrm{CO}\left(\mathrm{NH}_3\right)_6\right]^{3+}(\mathrm{aq})$ is $2.5 \times 10^6$. Calculate the value of $\Delta \mathrm{G}^{\circ}$ at $25^{\circ} \mathrm{C}\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$. In which direction is the reaction spontaneous under standard conditions?
Answer$\Delta\text{G}^\circ=-2.303\text{ RT}\log\text{K}$$=-2.303\times8.314\times298\log2.5\times10^6$
$=-5705.8[0.3980+6.0000]$ $[\log2.5=0.3980,\log10^6=6.0000]$
$=\frac{-5705.8\times6.3980}{1000}=-36.505\text{ kJ mol}^{-1}$
The reaction is spontaneous is forward direction under standard conditions because $\Delta\text{G}=-\text{ve}.$
View full question & answer→Question 462 Marks
For an isolated system, $\Delta\text{U}=0,$ what will be $\Delta\text{S}?$
AnswerChange in internal energy $(\Delta\text{U})$ for an isolated system is zero for it does not exchange any energy with the surroundings. But entropy tends to increase in case of spontaneous reaction. Therefore, $\Delta\text{S}>0$ or positive.
View full question & answer→Question 472 Marks
What is bond energy? Why is it called enthalpy of atomisation?
AnswerBond energy is the amount of energy released when bonds are formed between isolated atoms in gaseous state to form one mole of gaseous molecule. It is called enthalpy of atomisation because it may also be defined as the amount of energy required to dissociate bonds present between the atoms of 1 mole of a gaseous molecule into constituting atoms.
View full question & answer→Question 482 Marks
$\Delta\text{H}$ and $\Delta\text{S}$ for a reaction are found to be $-10000 \mathrm{~J} \mathrm{~mol}^{-1}$ and $-33.3 \mathrm{JK}^{-1}$. Under what conditions reaction will proceed in reverse direction.
Answer$\Delta\text{G}=\Delta\text{H}-\text{T}\Delta\text{S}$$0=-10000-\text{T}\times(-33.3)\text{J}$
$\Rightarrow\text{T}=\frac{10000}{33.3}=300.3\text{K}$
The reaction will proceed in backward direction above 300.3K because $\Delta\text{G}$ will +ve for forward reaction above 300.3K, hence $\Delta\text{G}$ will -ve for backward reaction.
View full question & answer→Question 492 Marks
Derive the relationship between isothermal and free expansion of an ideal gas.
AnswerWork done in isothermal reversible expansion of an ideal gas $\text{w}=-2.303\text{n RT}\log\frac{\text{V}_2}{\text{V}_1}$$=-2.303\text{n RT}\log\frac{\text{P}_1}{\text{P}_2}$
In free expansion of an ideal gas. w = 0 because ideal gases have negligible force of attraction, therefore. work done is zero in free expansion because no external force is acting.$\text{w}=-\text{P}_\text{ext}\Delta\text{V}$
$\text{P}_\text{ext}=0$
$\text{w}=0$
View full question & answer→Question 502 Marks
Calculate the total entropy change, $\Delta \mathrm{S}_{\text {total }}$ and state if the process is spontaneous, when one mole of liquid mercury $\mathrm{Hg}(\mathrm{l})$ changes to mercury vapour, $\mathrm{Hg}(\mathrm{g})$ at 298 K . The molar entropy of vapourisation of Hg is $99 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ and molar enthalpy of vapourisation is $59.1 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}$.
Answer$\Delta\text{S}_\text{surr}=\frac{-\Delta\text{H}_\text{sys}}{\text{T}}=\frac{-59100\text{J mol}^{-1}}{298\text{K}}$$=-198\text{JK}^{-1}\text{mol}^{-1}$
$\Delta\text{S}_\text{total}=\Delta\text{S}_\text{sys}+\Delta\text{S}_\text{surr}$
$=99\text{J}-198\text{J}=-99\text{JK}^{-1}\text{mol}^{-1}$
Since $\Delta\text{S}_\text{total}=-\text{ve},$ therefore, process is non-spontaneous at 298K.
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