NCERT Solution Class 11th Chemistry Chapter – 6 Thermodynamics Question & Answer

June 19, 2022

NCERT Solution Class 11th Chemistry Chapter – 6 Thermodynamics

Textbook	NCERT
class	Class – 11th
Subject	Chemistry
Chapter	Chapter – 6
Chapter Name	Thermodynamics
Category	Class 11th Chemistry Question & Answer
Medium	English
Source	last doubt

NCERT Solution Class 11th Chemistry Chapter – 6 Thermodynamics

?Chapter – 6?

✍Thermodynamics✍

?Question & Answer?

Q-1: Choose the correct answer. A thermodynamic state function is a quantity
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only

Answer: (ii) A quantity which is independent of path.

Reason: Functions like pressure, volume and temperature depends on the state of the system only and not on the path.

Q-2: For the process to occur under adiabatic conditions, the correct condition is:
(i) ∆T = 0 (ii) ∆p = 0
(iii) q = 0 (iv) w = 0

Answer: (iii) q = 0

Reason: For an adiabatic process heat transfer is zero, i.e. q = 0.

Q-3: The enthalpies of all elements in their standard states are:
(i) Unity
(ii) Zero
(iii) < 0
(iv) Different for every element

Answer: (ii) Zero

Q-4: ∆U0 of combustion of methane is – X kJ mol–1. The value of ∆H0 is
(i) = \Delta U^{\Theta }ΔUΘ
(ii) > \Delta U^{\Theta }ΔUΘ
(iii) < \Delta U^{\Theta }ΔUΘ
(iv) 0

Answer: (iii) < $\Delta U^{\Theta }$

Reason:

$\Delta H^{\Theta } = \Delta U^{\Theta } + \Delta n_{g}RT$ ; $\Delta U^{\Theta }$ = – Y $mol^{-1}$ ,

$\Delta H^{\Theta } = ( – Y) + \Delta n_{g}RT \Rightarrow \Delta H^{\Theta } < \Delta U^{\Theta }$

Q-5: The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy of formation of CH4(g) will be
(i) -74.8 kJmol^{-1}mol−1
(ii) -52.27 kJmol^{-1}mol−1
(iii) +74.8 kJmol^{-1}mol−1
(iv) +52 kJmol^{-1}mol−1

Answer: (i) -74.8kJ $mol^{-1}$

1. CH_4(g) + 2O_2(g) $\rightarrow$ CO_2(g) + 2H₂O_(g)

$\Delta H = -890.3kJmol^{-1}$

2. C_(s) + O_2(g) $\rightarrow$ CO_2(g)

$\Delta H = -393.5kJmol^{-1}$

3. 2H_2(g) + O_2(g) $\rightarrow$ 2H₂O_(g)

$\Delta H = -285.8kJmol^{-1}$

C_(s) + 2H_2(g) $\rightarrow$ CH_4(g)

$\Delta _{f}H_{CH_{4}}$ = $\Delta _{c}H_{c}$ + $2\Delta _{f}H_{H_{2}}$ – $\Delta _{f}H_{CO_{2}}$

= [ -393.5 +2(-285.8) – (-890.3)] kJ $mol^{-1}$

= -74.8kJ $mol^{-1}$

Q-6: A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be
(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(iv) possible at any temperature

Answer: (iv) possible at any temperature
\Delta GΔG should be –ve, for spontaneous reaction to occur
\Delta GΔG = \Delta HΔH – T\Delta SΔS
As per given in question,
\Delta HΔH is –ve ( as heat is evolved)
\Delta SΔS is +ve
Therefore, \Delta GΔG is negative
So, the reaction will be possible at any temperature.

Q-7: In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

Answer:

As per Thermodynamics 1^st law,
\Delta UΔU = q + W(i);
\Delta UΔU internal energy = heat
W = work done
W = -594 J (work done by system)
q = +701 J (+ve as heat is absorbed)
Now,
\Delta UΔU = 701 + (-594)
\Delta UΔU = 307 J

Q-8: The reaction of cyanamide, NH2CN(s), with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7 kJ mol–1 at 298 K. Calculate enthalpy change for the reaction at 298 K.
NH2CN(g) + 3/2 O2(g) → N2(g) + CO2(g) + H2O(l)

Answer:

\Delta HΔH is given by,
\Delta H = \Delta U + \Delta n_{g}RTΔH=ΔU+ΔngRT………………(1)
\Delta n_{g}Δng = change in number of moles
\Delta UΔU = change in internal energy
Here,
\Delta n_{g} = \sum n_{g}(product) – \sum n_{g}(reactant)Δng=∑ng(product)–∑ng(reactant)
= (2 – 1.5) moles
\Delta n_{g}Δng = 0.5 moles
Here,
T =298K
\Delta UΔU = -742.7 kJmol^{-1}kJmol−1
R = 8.314\times 10^{-3}kJmol^{-1}K^{-1}8.314×10−3kJmol−1K−1
Now, from (1)
\Delta H = (-742.7 kJmol^{-1}) + (0.5mol)(298K)( 8.314\times 10^{-3}kJmol^{-1}K^{-1})ΔH=(−742.7kJmol−1)+(0.5mol)(298K)(8.314×10−3kJmol−1K−1)
= -742.7 + 1.2
\Delta HΔH = -741.5 kJmol^{-1}kJmol−1

Q-9: Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol^–1 K^–1.

Answer:

Expression of heat(q),
q = mCP\Delta Tq=mCPΔT;………………….(a)
\Delta TΔT = Change in temperature
c = molar heat capacity
m = mass of substance
From (a)
q = ( \frac{60}{27}mol )(24mol^{-1}K^{-1})(20K)q=(2760mol)(24mol−1K−1)(20K)
q = 1066.67 J = 1.067 KJ

Q-10: Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C. ∆fusH = 6.03 kJ mol–1 at 0°C.
C_{p}[H_{2}O_{(l)}] = 75.3 J\;mol^{-1}K^{-1}Cp[H2O(l)]=75.3Jmol−1K−1
C_{p}[H_{2}O_{(s)}] = 36.8 J\;mol^{-1}K^{-1}Cp[H2O(s)]=36.8Jmol−1K−1

Answer: $\Delta H_{total}$ = sum of the changes given below:

(a) Energy change that occurs during transformation of 1 mole of water from $10^{\circ}C\;to\;0^{\circ}C$ .

(b) Energy change that occurs during transformation of 1 mole of water at $0^{\circ}C$ to 1 mole of ice at $0^{\circ}C$ .

(c) Energy change that occurs during transformation of 1 mole of ice from $0^{\circ}C\;to\;(-10)^{\circ}C$ .

\Delta H_{total} = C_{p}[H_{2}OCl]\Delta T + \Delta H_{freezing} C_{p}[H_{2}O_{l}]\Delta TΔHtotal=Cp[H2OCl]ΔT+ΔHfreezingCp[H2Ol]ΔT
= (75.3 J mol^{-1}K^{-1}Jmol−1K−1)(0 – 10)K + (-6.03*1000 J mol^{-1}Jmol−1(-10-0)K
= -753 J mol^{-1}Jmol−1 – 6030J mol^{-1}Jmol−1 – 368J mol^{-1}Jmol−1
= -7151 J mol^{-1}Jmol−1
= -7.151kJ mol^{-1}kJmol−1

Thus, the required change in enthalpy for given transformation is -7.151 $mol^{-1}$ .

Q-11 Enthalpy of combustion of carbon to CO₂ is –393.5 kJ mol^–1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.

Answer: Formation of carbon dioxide from di-oxygen and carbon gas is given as:

C_(s) + O_2(g) → CO_2(g); $\Delta _{f}HΔfH$ = -393.5 $kJ mol^{-1}kJmol−1$

1 mole CO₂ = 44g

Heat released during formation of 44g CO₂ = -393.5 $kJ mol^{-1}kJmol−1$

Therefore, heat released during formation of 35.2g of CO₂ can be calculated as

= $\frac{-393.5kJmol^{-1}}{44g}\times 35.2g44g−393.5kJmol−1×35.2g$

= -314.8 $kJ mol^{-1}kJmol−1$

Q-12: Enthalpies of formation of CO (g), CO2 (g), N2O (g) and N2O4(g) are –110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of ∆rH for the reaction:
N2O4(g) + 3CO(g)→ N2O(g) + 3 CO2(g)

Answer:

“ $\Delta _{r}H$ for any reaction is defined as the fifference between $\Delta _{f}H$ value of products and $\Delta _{f}H$ value of reactants.”

$\Delta _{r}H = \sum \Delta _{f}H (products) – \sum \Delta _{f}H (reactants)$

Now, for

N₂O_4(g) + 3CO_(g) à N₂O_(g) + 3 CO_2(g)

$\Delta _{r}H = [(\Delta _{f}H(N_{2}O) + (3\Delta _{f}H(CO_{2})) – (\Delta _{f}H(N_{2}O_{4}) + 3\Delta _{f}H(CO))]$

Now, substituting the given values in the above equation, we get:

$\Delta _{r}H$ = [{81 $mol^{-1}$ + 3(-393) $mol^{-1}$ } – {9.7 $mol^{-1}$ + 3(-110) $mol^{-1}$ }] $\Delta _{r}H$ = -777.7 $mol^{-1}$

Q-13: Given N₂ (g) + 3H₂ (g) → 2NH₃ (g) ; ∆_rH⁰= –92.4 kJ mol^–1
What is the standard enthalpy of formation of NH₃ gas?

Answer: “Standard enthalpy of formation of a compound is the 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.”

Dividing the chemical equation given in the question by 2, we get

(0.5)N_2(g) + (1.5)H_2(g) → 2NH_3(g)

Therefore, Standard Enthalpy for formation of ammonia gas
= (0.5) \Delta _{r}H^{\Theta }ΔrHΘ
= (0.5)(-92.4kJ mol^{-1}kJmol−1)
= -46.2kJ mol^{-1}kJmol−1

Q-14: Calculate the standard enthalpy of formation of CH₃OH_(l) from the following data:

CH₃OH_(l) + 3/2 O_2(g)→ CO_2(g) + 2H₂O_(l); \Delta _{r}H^{\Theta }ΔrHΘ = -726 kJ mol^{-1}kJmol−1

C_(g) + O_2(g)→ CO_2(g); \Delta _{c}H_{\Theta }ΔcHΘ = -393 kJ mol^{-1}kJmol−1

H_2(g) + 1/2 O_2(g)→ H₂O_(l); \Delta _{f}H^{\Theta }ΔfHΘ = -286 kJ mol^{-1}kJmol−1

Answer:

C_(s) + 2H₂O_(g) + (1/2)O_2(g) → CH₃OH_(l) …………………………(i)

CH₃OH_(l) can be obtained as follows,

$\Delta _{f}H_{\Theta }ΔfHΘ$ [CH₃OH_(l)] = $\Delta _{c}H_{\Theta }ΔcHΘ$

2 $\Delta _{f}H_{\Theta }ΔfHΘ$ – $\Delta _{r}H_{\Theta }ΔrHΘ$

= (-393 $kJ mol^{-1}kJmol−1$ ) +2(-286 $kJ mol^{-1}kJmol−1$ ) – (-726 $kJ mol^{-1}kJmol−1$ )

= (-393 – 572 + 726) $kJ mol^{-1}kJmol−1$

= -239 $kJ mol^{-1}kJmol−1$

Thus, $\Delta _{f}H_{\Theta }ΔfHΘ$ [CH₃OH_(l)] = -239 $kJ mol^{-1}kJmol−1$

Q-15: Calculate the enthalpy change for the process

CCl_4(g)→ C_(g) + 4Cl_(g) and determine the value of bond enthalpy for C-Cl in CCl_4(g).

$\Delta _{vap}H^{\Theta }ΔvapHΘ$ (CCl₄) = 30.5 $kJ mol^{-1}kJmol−1$ .

$\Delta _{f}H^{\Theta }ΔfHΘ$ (CCl₄) = -135.5 $kJ mol^{-1}kJmol−1$ .

$\Delta _{a}H^{\Theta }ΔaHΘ$ (C) = 715 $kJ mol^{-1}kJmol−1$ ,

$\Delta _{a}H^{\Theta }ΔaHΘ$ is a enthalpy of atomisation

$\Delta _{a}H^{\Theta }ΔaHΘ$ (Cl₂) = 242 $kJ mol^{-1}kJmol−1$ .

Answer:

“ The chemical equations implying to the given values of enthalpies” are:

(1) CCl_4(l) à CCl_4(g) ; $\Delta _{vap}H^{\Theta }$ = 30.5 $mol^{-1}$

(2) C_(s) à C_(g) $\Delta _{a}H^{\Theta }$ = 715 $mol^{-1}$

(3) Cl_2(g) à 2Cl_(g) ; $\Delta _{a}H^{\Theta }$ = 242 $mol^{-1}$

(4) C_(g) + 4Cl_(g) à CCl4(g); $\Delta _{f}H^{\Theta }$ = -135.5 $mol^{-1}$ $\Delta H$ for the process CCl_4(g) à C_(g) + 4Cl_(g)can be measured as:

$\Delta H = \Delta _{a}H^{\Theta }(C) + 2\Delta _{a}H^{\Theta }(Cl_{2}) – \Delta _{vap}H^{\Theta } – \Delta _{f}H$

= (715 $mol^{-1}$ ) + 2( $mol^{-1}$ ) – (30.5 $mol^{-1}$ ) – (-135.5 $mol^{-1}$ )

Therefore, $1304kJmol^{-1}$

The value of bond enthalpy for C-Cl in CCl_4(g)

= $\frac{1304}{4}kJmol^{-1}$

= 326 $mol^{-1}$

Q-16: For an isolated system, ∆U = 0, what will be ∆S?

Answer:

$\Delta U$ is positive ; $\Delta U$ > 0.

As, $\Delta U$ = 0 then $\Delta S$ will be +ve, as a result reaction will be spontaneous.

Q-17: For the reaction at 298K,
2A + B → C
\Delta HΔH = 400 kJ mol^{-1}kJmol−1
\Delta HΔH = 0.2 kJ mol^{-1}K^{-1}kJmol−1K−1
At what temperature will the reaction become spontaneous considering \Delta SΔS and \Delta HΔH to be constant over the temperature range?

Answer:

Now,
\Delta G = \Delta H – T\Delta SΔG=ΔH–TΔS
Let, the given reaction is at equilibrium, then \Delta TΔT will be:
T = (\Delta H – \Delta G)\frac{1}{\Delta S}(ΔH–ΔG)ΔS1 \frac{\Delta H}{\Delta S}ΔSΔH; (\Delta GΔG = 0 at equilibrium)
= 400kJ mol^{-1}kJmol−1/0.2kJ mol^{-1}K^{-1}kJmol−1K−1
Therefore, T = 2000K
Thus, for the spontaneous, \Delta GΔG must be –ve and T > 2000K.

Q-18: For the reaction

2Cl_(g)→ Cl_2(g)

What are the signs of $\Delta SΔS$ and $\Delta HΔH$ ?

Answer:

\Delta SΔS and \Delta HΔH are having negative sign.
The reaction given in the question represents the formation of Cl molecule from Cl atoms. As the formation of bond takes place in the given reaction. So, energy is released. So, \Delta HΔH is negative.
Also, 2 moles of Chlorine atoms are having more randomness than 1 mole of chlorine molecule. So, the spontaneity is decreased. Thus, \Delta SΔS is negative.

Q-19: For the reaction

2A_(g) + B_(g)→ 2D_(g)

$\Delta U^{\Theta }ΔUΘ$ = -10.5 kJ and $\Delta S^{\Theta }ΔSΘ$ = -44.1 $JK^{-1}JK−1$

Calculate $\Delta G^{\Theta }ΔGΘ$ for the reaction, and predict whether the reaction may occur spontaneously.

Answer:

2A_(g) + B_(g) → 2D_(g)
\Delta n_{g}Δng = 2 – 3
= -1 mole
Putting value of \Delta U^{\Theta }ΔUΘ in expression of \Delta HΔH:
\Delta H^{\Theta } = \Delta U^{\Theta } + \Delta n_{g}RTΔHΘ=ΔUΘ+ΔngRT
= (-10.5KJ) – (-1)( 8.314\times 10^{-3}kJK^{-1}mol^{-1}8.314×10−3kJK−1mol−1)(298K)
= -10.5kJ -2.48kJ
\Delta H^{\Theta }ΔHΘ = -12.98kJ
Putting value of \Delta S^{\Theta }ΔSΘ and \Delta H^{\Theta }ΔHΘ in expression of \Delta G^{\Theta }ΔGΘ:
\Delta G^{\Theta } = \Delta H^{\Theta } – T\Delta S^{\Theta }ΔGΘ=ΔHΘ–TΔSΘ
= -12.98kJ –(298K)(-44.1JK^{-1}JK−1)
= -12.98kJ +13.14kJ
\Delta G^{\Theta }ΔGΘ = 0.16kJ
As, \Delta G^{\Theta }ΔGΘ is positive, the reaction won’t occur spontaneously.

Q-20: The equilibrium constant for a reaction is 10. What will be the value of ∆G₀? R = 8.314 JK^–1 mol^–1, T = 300 K.

Answer:

Now,
\Delta G^{\Theta }ΔGΘ = -2.303RT\ln k−2.303RTlnk
= (2.303)( 8.314\times kJK^{-1}mol^{-1}8.314×kJK−1mol−1)(300K) \log 10log10
= -5527Jmol^{-1}Jmol−1
= -5.527kJmol^{-1}kJmol−1

Q-21: Comment on the thermodynamic stability of NO(g), given,

(1/2)N_2(g) + (1/2)O_2(g)→ NO_(g); \Delta _{r}H^{\Theta } = 90kJmol^{-1}ΔrHΘ=90kJmol−1

NO_(g) + (1/2)O_2(g) → NO_2(g); \Delta _{r}H^{\Theta } = -74kJmol^{-1}ΔrHΘ=−74kJmol−1

Answer:

• The +ve value of $\Delta _{r}H$ represents that during NO_(g) formation from O₂ and N₂, heat is absorbed. The obtained product, NO_(g) is having more energy than reactants. Thus, NO_(g) is unstable.

• The -ve value of $\Delta _{r}H$ represents that during NO_2(g) formation from O_2(g) and NO_(g), heat is evolved. The obtained product, NO_2(g) gets stabilized with minimum energy.

• Thus, unstable NO_(g)converts into stable NO_2(g).

Q-22: Calculate the entropy change in surroundings when 1.00 mol of H₂O(l) is formed under standard conditions. ∆_f H⁰= –286 kJ mol^–1.

Answer:

\Delta _{r}H^{\Theta } = -286kJmol^{-1}ΔrHΘ=−286kJmol−1 is given so that amount of heat is evolved during the formation of 1 mole of H₂O_(l).
Thus, the same heat will be absorbed by surrounding Qsurr = +286kJmol^{-1}kJmol−1.
Now, \Delta S_{surr}ΔSsurr = Qsurr/⁷
= \frac{286kJmol^{-1}}{298K}298K286kJmol−1
Therefore, \Delta S_{surr} = 959.73Jmol^{-1}K^{-1}ΔSsurr=959.73Jmol−1K−1