3 Marks Question - Chemistry STD 11 Science Questions - Vidyadip

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3 Marks Question

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54 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks

Calculate the lattice enthalpy of MgBr₂, given that
Enthalpy of formation of MgBr₂ = -524kJ mol^-1
Sublimation energy of Mg = 148kJ mol^-1
Ionization energy of Mg = 2187kJ mol^-1
Vaporisation energy of Br₂(l) = 31kJ mol^-1
Dissociation energy of Br₂(g) = 193kJ mol^-1
Electron gain enthalpy of Br₂(g) = 331kJ mol^-1

Answer

$\Delta_\text{f}\text{H}=\Delta\text{H}_\text{sub}+\Delta\text{H}_\text{ion}+\Delta\text{H}_\text{EA}+\Delta\text{H}_\text{D}\\+\Delta\text{H}_\text{Lattice}+\Delta\text{H}_\text{vap}$

$-524=148+2187-331+193\\+31+\Delta\text{H}_\text{Lattice}$

$\Delta\text{H}_\text{Lattice}=-524-2228=-2752\text{kJ mol}^{-1}$

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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.

Answer

Heat 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.

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Question 33 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 $\Delta\text{G}.$

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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.

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Question 53 Marks

Use the following data to calculate the $\Delta\text{G}^{\circ}$ for the reaction in which PF₅(g) forms from PF₃(g) and F₂(g) at 298K.

PF₃(g) + F₂(g) → PF₅(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

$\Delta \text{H}^{\circ}= \sum\Delta \text{H}^{\circ}_f \text{(products)} - \sum\Delta \text{H}^{\circ}_f \text{(reactants)}$

$= -1577 - (919)$

$= -658 \text{kJ}$

$\Delta \text{H}^{\circ}= \sum\text{S}^{\circ}_\text{products} - \sum \text{S}^{\circ}_\text{reactants}$

$= 301 - 273 - 203 = 175 \text{JK}^{-1}$

$= -0.175 \text{kJ} \text{ K}^{-1}$

$\Delta \text{G}^{\circ} = \Delta \text{H}^{\circ} - \text{T}\Delta \text{S}^{\circ}$

$= -658 \text{kJ} - 298 \times 0.175 \text{kJ}$

$= -606 \text{kJ}$

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Question 63 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.

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Question 73 Marks

Calculate the entropy change in surroundings when 1.00 mol of H₂O(l) is formed under standard conditions. $ \Delta_{\text{f}}\text{H}^\ominus=-286\text{kJ} \ \text{mol}^{-1}.$

Answer

It is given that 286kJ mol^–1 of heat is evolved on the formation of 1 mol of H₂O_(l). Thus, an equal amount of heat will be absorbed by the surroundings.

q_surr = +286kJ mol^–1

Entropy change $\Delta\text{S}_{\text{surr}}$ for the surroundings $=\frac{\text{q}_{\text{surr}}}{7}$

$=\frac{286\text{kJ} \ \text{mol}^{-1}}{298\text{k}}$

$\therefore\Delta\text{S}_{\text{surr}}=959.73\text{J} \ \text{mol}^{-1} \ \text{k}^{-1}$

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Question 83 Marks

1 mole of an ideal gas undergoes reversible isothermal expansion from an initial volume of V₁ to a final volume of 10V₁ and does 10kJ of work. The initial pressure was 1 × 10⁷Pa.

Calculate V₁.
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, p₁V₁ = n₁RT,

i.e., (10⁷Pa)V₁ = 1 × 8.314 × 522.3

V₁ = 4.342 × 10^-4m³

= 4.342 × 10²cm³

= 434.2cm³

We cannot apply the formula $-\text{W}=\text{p}\Delta\text{V}$ because expansion is not against constant pressure.

If there were 2 moles of the gas, applying p₁V₁ = n₁RT, we get (10⁷Pa)(4.342 × 10^-4m³) = 2 × 8.314T or T = 261.1K, i.e. half of the first value.

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Question 93 Marks

Enthalpy of combustion of carbon to CO₂ is –393.5kJ mol^–1. Calculate the heat released upon formation of 35.2g of CO₂ from carbon and dioxygen gas.

Answer

Formation of CO₂ from carbon and dioxygen gas can be represented as:

$\text{C}_{(\text{s})}+\text{O}_{{2(\text{g})}}\xrightarrow[]{\ \ \ \ \ \ \ }\text{CO}_{2(\text{g})} \ \Delta_{\text{f}}\text{H}=-393.5\text{kJ} \ \text{mol}^{-1}$

Heat released on formation of 44g CO₂ = –393.5kJ mol^–1

Heat released on formation of 35.2g CO₂

$=\frac{-393.5\text{kJ} \ \text{mol}^{-1}}{44\text{g}}\times35.2\text{g}$

$= –314.8\text{kJ} \ \text{mol}^{–1} $

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Question 103 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.

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Question 113 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.

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Question 123 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,

p₁V₁ = p₂V₂ or pV = constant

$\Delta\text{H}=\Delta\text{E}+\Delta(\text{pV})=0+0=0$

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Question 133 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

$40.15\times10^3\text{J}=\text{M}\times9.81\times\text{ms}^{-1}\times1\text{m}$

$\text{M}=\frac{40.15\times10^3\text{kg m}^2\text{s}^{-2}}{9.81\text{m s}^{-2}\times1\text{m}}$ $\big(\because\text{J}=\text{kg m}^2\text{s}^{-2}\big)$

$=4092.76\text{kg}$

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Question 143 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 × Molar mass of C₂H₅OH

$= 2.46\text{J} \text{ g}^{-1} \text{K}^{-1} \times 46 \text{g} \text{ mol}^{-1}$

$= 113.16 \text{J K}^{-1} \text{mol}^{-1}$

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Question 153 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

$\Delta\text{G}=\Delta\text{H}-\text{T}\Delta\text{S}$

$\Delta\text{G}=77.5\times1000\text{ J}-400\text{K}\times135\text{J K}^{-1}$

$=77500\text{J}-54000\text{J}=23500\text{J}$

$\Delta\text{G}=-2.303\text{RT}\log\text{K}$

$23500\text{J}=-2.303\times8.314\times400\text{K}\log\text{K}$

$\log\text{K}=\frac{-23500}{19.147\times400\text{K}}=\frac{-235}{76.588}$

$=-3.068+1-1=\bar{4}.932$

$\text{K}=\bar{4}.932,\text{ Antilog}=8.551\times10^{-4}$

$\text{Antilog }0.932=8.551$

$\text{Antilog }\bar{4}.932=8.551\times10^{-4}$

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Question 163 Marks

For the reaction,
2Cl(g) → Cl₂(g), what are the signs of $\Delta\text{H}$ and $\Delta\text{S}?$

Answer

$\Delta\text{H}$ and $\Delta\text{S}$ are negative

The 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.

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Question 173 Marks

How is the third law of thermodynamics useful in calculation of the absolute entropies? Calculate the value of $\Delta\text{S}^\circ$ for the following reaction at 400K:
2NOCl(g) → 2NO(g) + Cl₂(g)
If the value of equilibrium constant for the reaction at 400K is 1.958 × 10^-4 and $\Delta\text{H}^\circ=77.2\text{kJ mol}^{-1}$ (R = 8.314J K^-1mol^-1).

Answer

$\Delta\text{S}=\text{S}_\text{T}-\text{S}_0$

Where S_T is entropy at temperature TK

But from third law, S₀ = 0

Where S₀ is entropy at 0K

$\therefore\text{S}_\text{T}=\Delta\text{S}$

If we measure the energy required to raise the temperature of a crystalline substance from 0K to 298K, we can determine entropy change. The entropy S_T at 298K is called absolute entropy.

$\Delta\text{G}^\circ=-2.303\text{RT}\log\text{K}$

$=-2.303\times8.314\text{JK}^{-1}\text{mol}^{-1}\\\times400\text{K}\log1.958\times10^{-4}$

$=(-19.147\times400\log1.958\times10^{-4})\text{J mol}^{-1}$

$=[-7658.8\times(\log1.958+\log10^{-4})]\text{J mol}^{-1}$

$=[-7658.8\times(0.2917-4.0000)]\text{J mol}^{-1}$

$\Delta\text{G}^\circ=-7658.8\times-3.7183=+28477.72\text{J mol}^{-1}$

$=28.48\text{kJ mol}^{-1}$

$\Delta\text{G}^\circ=\Delta\text{H}^\circ-\text{T}\Delta\text{S}^\circ$

$28.48\text{kJ mol}^{-1}=77.20\text{kJ mol}^{-1}-400\Delta\text{S}^\circ$

$-400\Delta\text{S}^\circ=-48.73\text{kJ mol}^{-1}$

$\Delta\text{S}^\circ=\frac{48.73\times1000\text{J mol}^{-1}}{400\text{K}}=\frac{487.3}{4}$

$=121.82\text{JK}^{-1}\text{mol}^{-1}$

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Question 183 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

$\text{w}=-2.303\text{nRT}\log\frac{\text{V}_2}{\text{V}_1}$

$\text{n}=0.75\text{mol},\ \text{R}=8.314\text{JK}^{-1}\text{mol}^{-1}\text{T}$

$\text{T}=27^\circ\text{C}+273=300\text{K}$

$\text{w}=-2.303\times0.75\times8.314\\\times300\text{K}\log\frac{25\text{L}}{15\text{L}}$

$\text{w}=-19.147\times\frac{3}{4}\times300[\log5-\log3]$

$\text{w}=\frac{-19.147\times900}{4}\times[0.6990-0.4771]$

$\text{w}=\frac{-19.147\times900\times0.2219}{4}$

$=\frac{-3823.84}{4}$

$=-955.96\text{J}$

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Question 193 Marks

The heat of combustion of C₂H₆ is -368.4kcal. Calculate heat of combustion of C₂H₄, heat of combustion of H₂ is 68.32kcal mol^-1. $\Delta\text{H}$ for the following reaction is -37.1kcal.

$\text{C}_2\text{H}_4(\text{g})+\text{H}_2(\text{g})\overrightarrow{\ \ \ \ \ \ }\ \text{C}_2\text{H}_6(\text{g})$

Answer

$\text{C}_2\text{H}_4(\text{g})+\text{H}_2(\text{g})\overrightarrow{\ \ \ \ \ \ }\ \text{C}_2\text{H}_6(\text{g});$

$\Delta\text{H}=-37.1\text{kcal}$

$\Delta_\text{c}\text{H}^{\ominus}\text{C}_2\text{H}_6=-368.4\text{kcal},$ $\Delta_\text{c}\text{H}^\ominus\text{C}_2\text{H}_4=?$

$\Delta_\text{c}\text{H}^\ominus\text{H}_2(\text{g})=-68.32\text{kcal}$

$\Delta\text{H}=\sum\Delta_\text{c}\text{H}^\ominus(\text{reactants})-\sum\Delta_\text{c}\text{H}^\ominus(\text{products})$

$\Delta\text{H}=\Delta_\text{c}\text{H}^\ominus\text{C}_2\text{H}_4+\Delta_\text{c}\text{H}^\ominus\text{H}_2(\text{g})\\-\Delta_\text{c}\text{H}^\ominus\text{C}_2\text{H}_6(\text{g})$

$-37.1\text{kcal}=\Delta_\text{c}\text{H}^\ominus\text{C}_2\text{H}_4-68.32-(-368.4)$

$\Delta_\text{c}\text{H}^\ominus\text{C}_2\text{H}_4=-337.18\text{kcal}$

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Question 203 Marks

Starting with the thermodynamic relationship G = H - TS, derive the following relationship: $\Delta\text{G}=-\text{T}\Delta\text{S}_\text{Total}$

Answer

$\text{G}=\text{H}-\text{TS},$ $\text{G}_1=\text{H}_1-\text{TS}_1,$ $\text{G}_2=\text{H}_2-\text{TS}_2$

$\text{G}_2-\text{G}_1=\text{H}_2-\text{H}_1-\text{T}(\text{S}_2-\text{S}_1)$

$\Delta\text{G}=\text{H}_2-\text{H}_1-\text{T}(\text{S}_2-\text{S}_1)$

$\Delta\text{G}=\Delta\text{H}-\text{T}\Delta\text{S}$

$\Delta\text{S}_\text{Total}=\Delta\text{S}_\text{sys}+\Delta\text{S}_\text{surr}$

$\Rightarrow\Delta\text{S}_\text{Total}=\Delta\text{S}_\text{sys}-\frac{\text{q}}{\text{T}}$

$\Delta\text{S}_\text{sys}=\frac{+\text{q}}{\text{T}},\Delta\text{S}_\text{surr}=\frac{-\text{q}}{\text{T}}$

$\Rightarrow\text{T}\Delta\text{S}_\text{Total}=\text{T}\Delta\text{S}_\text{sys}-\text{q}=\text{T}\Delta\text{S}_\text{sys}-\Delta\text{H}$

$\text{T}\Delta\text{S}_\text{Total}=-\Delta\text{G}$

$\Rightarrow\Delta\text{G}=-\text{T}\Delta\text{S}_\text{Total}$

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Question 213 Marks

$\text{X(g)}+3\text{Y(g)}\rightleftharpoons2\text{Z(g)};$ $\Delta\text{H}=-40\text{kJ}$ and s^o of X, Y and Z are 60, 40 and 50kJ^-1 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.

$\therefore\Delta\text{H}>\text{T}\Delta\text{S}$

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Question 223 Marks

Using the data (all values are in kilocalories per mole at 25°C) given below, calculate the bond energy of C-C and C-H bonds.

$\Delta\text{H}^\circ$ combustion (ethane) = -372.0

$\Delta\text{H}^\circ$ combustion (propane) = -530.0

$\Delta\text{H}^\circ$ for C(graphite) → C(g)= 172.0

Bond energy of H-H = 104.0

$\Delta_\text{f}\text{H}^\circ$ of H₂O(l) = -68.0

$\Delta\text{H}^\circ$ for CO₂(g) = -94.0

Answer

We 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 C₂H₆ (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 C₃H₈(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.

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Question 233 Marks

Calculate bond energy of C-H bond if $\Delta_\text{c}\text{H}$ of CH₄, is -891kJ mol^-1, $\Delta_\text{c}\text{H}$ of C(s) is -394kJ mol^-1, $\Delta_\text{c}\text{H}$ of H₂(g) is -286kJ mol^-1, heat of sublimation of C(s) is 717kJ mol^-1, heat of dissociation of H₂ is 436kJ 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 $=\frac{1664}{4}=416\text{kJ mol}^{-1}.$

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Question 243 Marks

Use the following data to calculate $\Delta_\text{lattice}\text{H}^\ominus$ for NaBr

$\Delta_\text{sub}\text{H}^\ominus$ for sodium metal = 108.4kJ mol^–1

Ionization enthalpy of sodium = 496kJ mol^–1

Electron gain enthalpy of bromine = –325kJ mol^–1

Bond dissociation enthalpy of bromine = 192kJ mol^–1

$\Delta_\text{f}\text{H}^\ominus$ for NaBr (s) = – 360.1kJ 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,

$\Delta_\text{f}\text{H}^\circ=\Delta_\text{sub}\text{H}^\circ+\frac{1}{2}\Delta\text{diss}\text{H}^\circ+\Delta_\text{i}\text{H}^\circ+\Delta_\text{eg}\text{H}^\circ+\Delta_\text{lattice}\text{H}^\circ$

$\therefore\ \Delta_\text{lattice}\text{H}^\circ=-360.1-108.4-96-496+325$

$=-735.5\text{kJ }\text{mol}^{-1}$

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Question 253 Marks

Calculate $\Delta_\text{f}\text{H}^\circ$ of HCl if bond energy of H-H bond is 437kJ mol^-1, Cl-Cl bond is 244kJ mol^-1 and H-Cl is 433kJ 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.

$=\frac{1}{2}\text{B}_{\text{H}-\text{H}}+\frac{1}{2}\text{B}_{\text{Cl}-\text{Cl}}-\text{B}_{\text{H}-\text{Cl}}$

$=218.5+122-433$

$=-92.5\text{kJ mol}^{-1}$

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Question 263 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.

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Question 273 Marks

The enthalpy of vaporisation of liquid diethyl ether (C₂H₅)₂O is 26.0kJ 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)

$\Delta_\text{cond}\text{S}^\circ=\frac{\Delta_\text{cond}\text{H}^\circ}{\text{T}}$

$=\frac{-26.0\times10^3\text{mol}^{-1}}{308}$

$=-84.4\text{JK}^{-1}\text{mol}^{-1}$

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Question 283 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, C₃H₈(g) is -103.8kJ 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

$3\text{C}_\text{(graphite)}+4\text{H}_2\text{O}\overrightarrow{\ \ \ \ \ \ \ }\ \text{C}_3\text{H}_8(\text{g})$

$\Delta_\text{r}\text{S}^\circ=\sum\text{S}^\circ_\text{m}(\text{products})-\text{S}^\circ_\text{m}(\text{reactants})$

$\Delta_\text{r}\text{S}^\circ=\text{S}^\circ_\text{m}[\text{C}_3\text{H}_8(\text{g})]-3\text{S}^\circ_\text{m}[\text{C(graphite)}]\\-4\text{S}^\circ_\text{m}[\text{H}_2]$

$=(270.2-3\times5.70-4\times130.7)\text{JK}^{-1}\text{mol}^{-1}$

$=(270.2-17.10-522.80)\text{JK}^{-1}\text{mol}^{-1}$

$=(270.2-539.90)=-269.7\text{JK}^{-1}\text{mol}^{-1}$

$\Delta_\text{r}\text{H}^\circ=\Delta_\text{f}\text{H}^\circ_\text{products}-\Delta_\text{f}\text{H}^\circ_\text{reactants}$

$\Delta_\text{r}\text{H}^\circ=\Delta_\text{f}\text{H}^\circ(\text{C}_3\text{H}_8)-4\Delta_\text{f}\text{H}^\circ[\text{H}_2(\text{g})]\\-3\Delta_\text{f}\text{H}^\circ[\text{C(graphipte)}]$

$\Delta_\text{r}\text{H}^\circ=-103.8\text{kJ mol}^{-1}$

$\Delta_\text{r}\text{G}^\circ=\Delta_\text{r}\text{H}^\circ-\text{T}\Delta_\text{r}\text{S}^\circ$

$=-103.8-\Big(\frac{298\times(-269.7)}{1000}\Big)\text{kJ mol}^{-1}$

$=(-103.80+80.370)\text{kJ mol}^{-1}$

$=-23.43\text{kJ mol}^{-1}$

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Question 293 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 [N₂H₄(l), hydrogen peroxide [H₂O₂(l)] and water [H₂O(l)] are -50.4, -193.2 and -242.7 kJ/ 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.
$\text{N}_2\text{H}_4\text{(l)}+2\text{H}_2\text{O}_2(\text{l})\overrightarrow{\ \ \ \ \ \ \ }\ \text{N}_2(\text{g})+4\text{H}_2\text{O(l)}$

$\Delta_\text{r}\text{H}=\Delta_\text{f}\text{H}(\text{N}_2)+4\Delta_\text{f}\text{H}(\text{H}_2\text{O})\\-\Delta_\text{f}\text{H}(\text{N}_2\text{H}_4)-2\Delta_\text{f}\text{H}(\text{H}_2\text{O})$

$=0+4\times[-242.7-(-50.4)-2(-193.2)]$

$=(-970.8+50.4+386.4)\text{kJ mol}^{-1}$

$\Delta_\text{r}\text{H}=-534\text{kJ mol}^{-1}$

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Question 303 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.

$\text{CO}_2(\text{g})+2\text{H}_2\text{O(l)}\overrightarrow{ \ \ \ \ \ \ }\ \text{CH}_4(\text{g})+2\text{O}_2(\text{g})$

$\Delta_\text{r}\text{H}=\Delta_\text{f}\text{H}(\text{CH}_4)+2\Delta_\text{f}\text{H}(\text{O}_2)-\Delta_\text{f}\text{H}(\text{CO}_2)\\-2\Delta_\text{f}\text{H}[\text{H}_2\text{O(l)}]$

$+890.3\text{kJ mol}^{-1}=\Delta_\text{f}\text{HCH}_4+2\times0-\\-(-393.51\text{kJ mol}^{-1})-2\times(-285.8\text{kJ mol}^{-1})$

$\Delta_\text{f}\text{H}(\text{CH}_4)=(+890.3-393.51\text{kJ}\\-571.6)\text{kJ mol}^{-1}$

$\Delta_\text{f}\text{H}(\text{CH}_4)=-74.8\text{kJ mol}^{-1}$

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Question 313 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, $\Delta_\text{f}\text{G}^\circ[\text{HC}\equiv\text{CH(g)}]=2.09\times10^5\text{J mol}^{-1},$ $\Delta_\text{f}\text{G}^\circ[\text{C}_6\text{H}_6(\text{g})]=1.24\times10^5\text{J mol}^{-1},$ R = 8.314JK^-1 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.

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Question 323 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

p₁ = 1bar, p₂ = 100bar, V₁ = 100mL, V₂ = 99mL

For adiabatic process, q = 0, $\Delta\text{U}=\text{q}+\text{W}$

$\Delta\text{U}=\text{W}$

$\text{W}=-\text{p}\Delta\text{V}$

$=-100(99-100)=100\text{bar mL}$

$\Delta\text{H}=\Delta\text{U}+\Delta\text{pV}$

$=100+\text{p}_2\text{V}_2-\text{p}_1\text{V}_1$

$=100+(100\times99)-(1\times100)$

$=100+9900-100=9900\text{bar mL}$

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Question 333 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 NH₃ gas?

Answer

Standard 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 NH_3(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 NH_3(g)

$=\frac{1}{2}\Delta_{\text{r}}\text{H}^\ominus$

$=\frac{1}{2}(-92.4\text{kJ}) \ \text{mol}^{-1}$

$=-46.2\text{kJ} \ \text{mol}^{-1}$

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Question 343 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.

Answer

From 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.

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Question 353 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 $\text{A}\rightleftharpoons\text{B}$ is 1.8 × 10^-7 at 298K. Calculate the value of $\Delta\text{G}^\circ$ for the reaction (R = 8.314JK^-1mol^-1). Predict the feasibility of the reaction under standard states.

Answer

According 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 × 10^-7, T = 298K, $\Delta\text{G}^\circ=?,$ R = 8.314JK^-1 mol^-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.

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Question 363 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 × 10⁵. [R = 8.314J K^-1mol^-1]

Answer

The 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.

$\Delta\text{G}^\circ=-2.303\text{RT}\log\text{K}$

$=-2.303\times8.314\text{JK}^{-1}\text{mol}^{-1}\\\times298\text{K}\log6.6\times10^5$

$=-19.147\text{J}\times298\log6.6\times10^5$

$=-5705.8[\log6.6+\log10^5]$

$=-5705.8[0.8195+5.0000]$

$=-5705.8\times5.8195\text{J}=-33204.903\text{J}$

$\Delta\text{G}^\circ=-33.205\text{kJ mol}^{-1}$

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Question 373 Marks

Explain the term 'standard molar free energy of formation’ of a compound. Calculate the equilibrium constant for the reaction:

$2\text{SO}_2(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{SO}_3(\text{g})$ at 25°C.

Given: $\Delta_\text{f}\text{G}^\circ\text{SO}_3(\text{g})=-371.1\text{kJ mol}^{-1},$ $\Delta_\text{f}\text{G}^\circ\text{SO}_2(\text{g})=-300.2\text{kJ mol}^{-1},$ R = 8.31JK^-1 mol^-1.

Answer

Standard free energy of formation of compound is defined as free energy change when a compound is formed from its constituting elements at 298K and 1atm pressure.

$2\text{SO}_2(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{SO}_3(\text{g})$

$\Delta\text{G}^\circ=\sum\Delta_\text{f}\text{G}^\circ(\text{products})-\sum\Delta_\text{f}\text{G}^\circ(\text{reactants})$

$=2\Delta_\text{f}\text{G}^\circ(\text{SO}_3)-2\Delta_\text{f}\text{G}^\circ(\text{SO}_2)-\Delta_\text{f}\text{G}^\circ(\text{O}_2)$

$=2\times(-371.1)-2\times(-300.2)-0$

$=-742.2+600.4=-141.8$

$\Delta\text{G}^\circ=-2.303\text{RT}\log\text{K}$

$-141.8\times1000\text{J}=-2.303\times8.314\times298\log\text{K}$

$\log\text{K}=\frac{1,41,800}{5705.8}=24.8519$

$\Rightarrow\text{K}=7.114\times10^{24}$

[Antilog of 0.8519 = 7.114]

[Antilog of 24.8519 = 7.114 × 10²⁴]

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Question 383 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 $\Delta\text{H}^\circ_\text{f}$ at 298K are NO(g) = 90.4kJ mol^-1 and NO₂(g) = 33.8kJ mol^-1, $\Delta\text{S}^\circ$ at 298K for the reaction = -70.0JK^-1mol^-1, R = 8.31JK^-1mol^-1.

Answer

Enthalpy 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, $\Delta\text{G}^\circ=\Delta\text{H}^\circ-\text{T}\Delta\text{S}^\circ,\ \text{T}=298\text{K}$

$\therefore\Delta\text{G}^\circ=-56600-298\times(-70)$

$=-35740\text{J mol}^{-1}$

$\log\text{K}=-\frac{\Delta\text{G}}{2.303\text{RT}}$

$\log\text{K}=-\frac{-35740}{2.303\times8.31\times298}=6.267$

$\therefore\text{K}=\text{anti}\log(6.267)=1.85\times10^6$

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Question 393 Marks

Calculate the standard Gibbs energy change for the formation of propane at 298K.

3C(graphite) + 4H₂(g) → C₃H₈(g)

$\Delta_\text{f}\text{H}^\circ$ for propane, C₃H₈(g) is -103.8kJ mol^-1

Given, $\text{S}^\circ_\text{m}\text{C}_3\text{H}_8(\text{g})=270.2\text{JK}^{-1}\text{mol}^{-1}$

$\text{S}^\circ_\text{m}\text{C(graphite)}=5.70\text{JK}^{-1}\text{mol}^{-1}$ and $\text{S}^\circ_\text{m}\text{H}_2(\text{g})=130.7\text{JK}^{-1}\text{mol}^{-1}$

Answer

3C(graphite) + 4H₂(g) → C₃H₈(g)

$\Delta_\text{r}\text{S}=\sum\text{S}^\circ_\text{m}(\text{products})-\text{S}^\circ_\text{m}(\text{reactants})$

$\Delta_\text{r}\text{S}=\text{S}^\circ_\text{m}[\text{C}_3\text{H}_8(\text{g})-3\text{S}^\circ_\text{m}[\text{C(graphite)}]\\+4\text{S}^\circ_\text{m}[\text{H}_2]]$

$=(270.2-3\times5.70-4\times130.7)\text{JK}^{-1}\text{mol}^{-1}$

$=(270.2-17.10-522.80)\text{JK}^{-1}\text{mol}^{-1}$

$=(270.2-539.90)=-269.7\text{JK}^{-1}\text{mol}^{-1}$

$\Delta_\text{r}\text{H}^\circ=\Delta_\text{f}\text{H}^\circ\text{products}-\Delta_\text{f}\text{H}^\circ\text{reactants}$

$\Delta_\text{r}\text{H}^\circ=\Delta_\text{f}\text{H}^\circ(\text{C}_3\text{H}_8)-4\Delta_\text{f}\text{H}^\circ[\text{H}_2(\text{g})]\\-3\Delta_\text{f}\text{H}^\circ[\text{C(graphite)}]$

$\Delta_\text{r}\text{H}^\circ=-103.8\text{kJ mol}^{-1};\ \Delta_\text{r}\text{G}^\circ=\Delta_\text{r}\text{H}^\circ-\text{T}\Delta_\text{r}\text{S}^\circ$

$=\Big(-103.8-\frac{298\times(-269.7)}{1000}\Big)\text{kJ mol}^{-1}$

$=(-103.80+80.370)\text{kJ mol}^{-1}$

$=-23.43\text{kJ mol}^{-1}$

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Question 403 Marks

What is meant by entropy? Predict the sign of entropy change in each of the following:

H₂ (at 298K, 1atm) → H₂(at 298K, 10atm)
H₂O (at 298K, 1atm) → H₂O (at 330K, 1atm)
2NH₄NO₃(s) → 2N₂(g) + 4H₂O(g) + O₂(8)
Crystallization of copper sulphate from its saturated solution.
$2\text{SO}_2(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{SO}_3(\text{g}).$

Answer

Entropy is defined as the degree of randomness or disorder.

$\Delta\text{S}=-\text{ve}$
$\Delta\text{S}=+\text{ve}$
$\Delta\text{S}=+\text{ve}$
$\Delta\text{S}=-\text{ve}$
$\Delta\text{S}=-\text{ve}$

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Question 413 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 NH₃ is -91.8kJ mol^-1. Calculate enthalpy change for the following reaction:

$2\text{NH}_3(\text{g})\rightarrow\text{N}_2(\text{g})+3\text{H}_2(\text{g})$

Answer

No, 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,

$\therefore\ \Delta_\text{r}\text{H}^\circ\neq\Delta_\text{f}\text{H}^\circ$

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Question 423 Marks

The combustion of one mole of benzene takes place at 298K and 1atm. After combustion, CO₂(g) and H₂O(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 $\Delta_\text{f}\text{H}^\circ\text{H}_2\text{O(l})=-285.83\text{kJ mol}^{-1}.$

Answer

Combustion 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$

$\Delta_\text{f}\text{H}^\circ\text{C}_6\text{H}_6(\text{l})=[6(-393.5)\\+3(-285.83)+3267]\text{kJ mol}^{-1}$

$=[-2361.0-857.49+3267.0]\text{kJ mol}^{-1}$

$=48.51\text{kJ mol}^{-1}$

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Question 433 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.

Answer

Here, n = 2 moles, T = 27°C = 300K, P₁ = 1atm, P₂ = 10atm

$\text{W}=2.303\text{nRT}\log\frac{\text{p}_2}{\text{p}_1}$

$=2.303\times2\times8.314\text{JK}^{-1}\text{mol}^{-1}\times300\text{K}\\\times\log\frac{10}{1}=11488.28\text{J}$

For isothermal compression of ideal gas, $\Delta\text{U}=0$

Further, $\Delta\text{U}=\text{q}+\text{W}$

$\therefore\text{q}=-\text{W}=-11488\text{J}$

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Question 443 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

$\therefore\Delta\text{U}=0+0=0$

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Question 453 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.

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Question 463 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)

Answer

Energy 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 $=\frac{32640}{2500}=13\text{days}$

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Question 473 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.

Answer

Heat 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.

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Question 483 Marks

Comment on the thermodynamic stability of NO(g), given

$\frac{1}{2}\text{N}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g})\xrightarrow{ \ \ \ \ \ \ } \text{NO}_{(\text{g})} \ ; \ \Delta_{\text{r}}\text{H}^\ominus=90\text{kJ} \ \text{mol}^{-1}$

$\text{NO}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g})\xrightarrow{ \ \ \ \ \ \ } \text{NO}_2(\text{g}) \ ; \ \Delta_{\text{r}}\text{H}^\ominus=-74\text{kJ} \ \text{mol}^{-1}$

Answer

The positive value of $\Delta_\text{r}\text{H}$ indicates that heat is absorbed during the formation of NO_(g). This means that NO_(g)has higher energy than the reactants (N₂ and O₂). Hence, NO_(g) is unstable.
The negative value of $\Delta_\text{r}\text{H}$ indicates that heat is evolved during the formation of NO_2(g) from NO_(g) and O_2(g). The product, NO_2(g) is stabilized with minimum energy.
Hence, unstable NO_(g) changes to stable NO_2(g).

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Question 493 Marks

Calculate the work done when 11.2g of iron dissolves in hydrochloric acid in:

A closed vessel.
An open beaker at 25°C (Atomic mass of Fe = 56u).

Answer

Iron reacts with hydrochloric acid to produce H, gas as

Fe(s) + 2HCl (aq) → FeCl₂(aq) + H₂(g)

Thus, 1 mole of Fe, i.e. 56g Fe produces H₂ gas = 1mol.

$\therefore$ 11.2g Fe will produce H₂ gas $=\frac{1}{56}\times11.2=0.2\text{mol}$

If the reaction is carried out in closed vessel, $\Delta\text{V}=0$

$\therefore\text{W}=-\text{p}_\text{ext}\Delta\text{V}=0$

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 H₂ at 25°C and 1atm pressure can be calculated as follows pV = nRT

$\therefore\text{V}=\frac{\text{nRT}}{\text{p}}$

$=\frac{0.2\text{mol}\times0.0821\text{L atm K}^{-1}\text{mol}^{-1}\times298\text{K}}{1\text{atm}}$

$=4.89\text{L}$

$\therefore\Delta\text{V}=\text{V}_\text{final}-\text{V}_\text{initial}=4.89\text{L}$

$\text{W}=-\text{p}_\text{ext}\Delta\text{V}=-1\text{atm}\times4.89\text{L}$

$=-4.89\text{L atm}=-4.89\times101.3\text{J}$

$=-495.4\text{J}$

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Question 503 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)

Answer

Work 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.

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Question 513 Marks

6.72dm³ of an unknown gas at STP requires 113.35J of heat to raise its temperature by 15°C at constant volume. Calculate C_V, C_p, and atomicity of the gas.

Answer

22.4dm³ of a gas at STP = 1mol

$\therefore$ 5.6dm³ of the gas at STP $=\frac{1}{22.4}\times6.72=0.3\text{mol}$

Thus, for 15° rise, 0.3mol of the gas at constant volume require heat = 113.35J

$\therefore$ For 1° rise, 1 mole of the gas at constant volume will require heat $=\frac{113.35}{15\times0.3}\text{J}=25.19\text{J}$

$\therefore\text{C}_\text{V}=25.19\text{JK}^{-1}\text{mol}^{-1}$

Now, $\text{C}_\text{p}=\text{C}_\text{V}+\text{R}$

$=25.19\text{JK}^{-1}\text{mol}^{-1}+8.314\text{JK}^{-1}\text{mol}^{-1}$

$=33.5\text{JK}^{-1}\text{mol}^{-1};$

$\therefore\gamma=\frac{\text{C}_\text{p}}{\text{C}_\text{V}}=\frac{33.5}{25.19}=1.33$

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Question 523 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?

Answer

According 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.

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Question 533 Marks

Calculate the number of kJ of heat necessary to raise the temperature of 60.0g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24J mol^–1 K^–1.

Answer

Mass 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}}$

= $60\times\frac{24}{27}\times20=\frac{28800}{27}=1066.67 \text{J}$

$= 1.068\text{kJ or 1.07kJ}$

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Question 543 Marks

Choose the correct answer.

$\Delta\text{U}^\ominus$of combustion of methane is – X kJ mol^–1. The value of $\Delta\text{H}^\ominus$ is:

$=\Delta\text{U}^\ominus$
$>\Delta\text{U}^\ominus$
$<\Delta\text{U}^\ominus$
$=0$

Answer

$<\Delta\text{U}^\ominus$

Explanation:

The balanced chemical equation fir the combustion reaction is:

$\text{CH}_{4(\text{g})}+2\text{O}_{{2(\text{g})}\xrightarrow[]{\ \ \ \ \ \ \ \ \ \ \ \ \ }\text{CO}_2(\text{g})}+2\text{H}_2\text{O}(\text{l})$

$\Delta_{\text{ng}}=1-3=-2$

$\Delta\text{H}^\ominus=\Delta\text{U}^\ominus+\Delta_\text{ng}\text{RT}=\Delta\text{U}^\ominus-2\text{RT}$

$\therefore \Delta\text{H}^\ominus<\Delta\text{U}^\ominus$

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