Class 11 - Chemistry - Thermodynamics

Question 6.1

Choose the correct answer. A thermodynamic state function is a quantity

  • Used to determine heat changes
  • Whose value is independent of path
  • Used to determine pressure volume work
  • Whose value depends on temperature only.

Answer.

Correct option is (ii).

Pressure, volume and temperature depends on the state of the system only, not on the path.

 

Question 6.2

For the process to occur under adiabatic conditions, the correct condition is

  • ΔT = 0
  • Δp = 0
  • Q = 0
  • W = 0

Answer.

Correct option is (iii)

For an adiabatic process, heat transfer is zero.

Question 6.3

The enthalpies of all elements in their standard states are

  • Unity
  • Zero
  • < 0
  • Different for each element

Answer.

Correct option is (ii)

 

 

Question 6.4

ΔUof combustion of methane is – X kJ mol-1. The value of ΔH is

  • = ΔU
  • > ΔU
  • < ΔU
  • = 0

Answer.

Correct option is (iii)

 

 

Question 6.5

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

  • – 74.8 kJ mol-1
  • -52.27 kJ mol-1
  • + 74.8 kJ mol-1
  • + 52.26 kJ mol-1

Answer.

Correct option is (i)

CH4 + 2O2 --> CO2 + 2H2O, Δc H = -890.3 kJ mol-1 –(i)

C + O2 --> CO2, Δc H = -393.5 kJ mol-1 –(ii)

H2 + ½ O2 --> H2O, Δc H = -285.8 kJ mol-1 –(iii)

We need to find out,

C + 2H2 --> CH4 , Δf H = ?

Eqn (ii) + 2 x (iii) – (i), then

= -393.5 + 2 x -285.8 – (-890.3) = -74.8 kJ mol-1

 

Question 6.6

A reaction A + B --> C + D + q is found to have a positive entropy change. The reaction will be

  • Possible at high temperature
  • Possible at low temperature
  • Not possible at any temperature
  • Possible at any temperature

Answer.

Correct option is (iv)

 ΔG should be –ve, for spontaneous reaction to occur

ΔG = ΔH –TΔS

As given,

ΔH is –ve ( as heat is evolved)

ΔS is +ve

Therefore, ΔG is negative

So, the reaction will be possible at any temperature.

 

Question 6.7

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.

Heat absorbed by the system, q = 701 J

Work done by the system = -394 J

Change in internal energy (ΔU = q + w = 701 – 394 = 307 J

 

 

 

Question 6.8

The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and ΔU 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.

ΔH = ΔU + ΔngRt ---(i)

Δng = change in number of moles

ΔU = change in internal energy

Δng = Σ ng (product) - Σ ng (reactant)

Δng = 2 – 1.5 moles = 0.5 moles

T = 298 K

ΔU = - 742.7 kJ mol-1

R = 8.314 x 10-3 kJ mol-1 K-1

Putting values in equation (i)

ΔH = (-742.7 kJ mol-1) + (0.5 mol) (298 K) (8.314 x 10-3 kJ mol-1K-1)

ΔH = -742.7 + 1.2 = -741.5 kJ mol-1

 

 

Question 6.9

Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminum from 35o C to 55o C.

Molar heat capacity of Al is 24 J mol-1 K-1.

Answer.

Number of moles of Al (m) = 60 g/(27 g mol-1) = 2.22 mol

Molar heat capacity (C) = 24 J mol-1K-1

Rise in temperature (ΔT) = 55 – 35 = 20 K

Heat evolved (q) = C x m x T = 24 x 2.22 x 20 = 1065.6 J

q = 1.067 kJ

 

 

Question 6.10

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0o C to ice at -10.0o C. ΔfusH = 6.03 kJ mol-1 at 0o C.

Cp [H2O (l)] = 75.3 J mol-1 K-1

Cp [H2O (s)] = 36.8 J mol-1 K-1.

Answer.

The above information can be represented as

According to Hess’s Law,  ΔH = ΔH1 + ΔH2 + ΔH3

ΔH1 = 75.3 J mol-1 K-1 (10 K) = 753 J mol-1

ΔH2 = -6.03 kJ mol-1 = -6030 J mol-1

ΔH3 = 36.8 J mol-1 K-1 (-10 K) = -368 J mol-1

ΔH = (753 – 6030 – 368) J mol-1 = -5645 J mol-1 = - 5.645 kJ mol-1

 

 

 

Question 6.11

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.

Equation will be as follows

C (s) + O2 (g) --> CO2 (g), ΔcH = -393.5 kJ mol-1

 

Heat released in the formation of 44 g of CO2 = 393.5 kJ

Heat released in the formation of 35.2 g of CO2

= (393.5 kJ) x (35.2g) /(44g)

= 314.8 kJ

 

 

Question 6.12

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 ΔHf for the reaction:

N2O4 (g) + 3CO --> 2 N2O (g) + 3CO2 (g)

Answer.

Enthalpy of reaction, ΔfH = [81 + 3 (-393)] – [9.7 + 3 ( -110)]

= -778 kJ mol-1

 

 

Question 6.13

Given

N2 (g) + 3 H2 (g) --> 2NH3 (g); ΔfH = - 92.4 kJ mol-1

What is the standard enthalpy of formation of NH3 gas?

Answer.

ΔH- NH3 (g) = - (92.4)/2 = -46.2 kJ mol-1

 

 

Question 6.14

Calculate the standard enthalpy of formation of CH3OH (l) from the following data.

CH3OH (l) + (3/2) O2 (g) --> CO2 (g) + 2H2O (l); ΔrH = -726 kJ mol-1

C (graphite) + O2 (g) --> CO2(g) ; ΔcH = - 393 kJ mol-1

H2 (g) + (½) O2 (g) --> H2O (l) , ΔfH = - 286 kJ mol-1

Answer.

CH3OH (l) + (3/2) O2 (g) --> CO2 (g) + 2H2O (l); ΔrH = -726 kJ mol-1 ----(i)

C (graphite) + O2 (g) --> CO2(g) ; ΔcH = - 393 kJ mol-1 ----(ii)

H2 (g) + (½) O2 (g) --> H2O (l) , ΔfH = - 286 kJ mol-1 (iii)

 

C (s) + 2 H2 (g) + ½ O2 (g) --> CH3OH (l) , ΔfH = ? ----(iv)

Multiply equation (iii) by 2 and then add to equation (ii)

C + 2H2 + 2O2 --> CO2 + 2H2O

ΔH = - (393 + 522) = - 965 kJ mol-1

Subtract equation (iv) from equation (i)

CH3OH + (3/2) O2 --> CO2 + 2H2O, ΔH = -726 kJ mol-1

Subtract,

C + 2H2 + (½) O2 --> CH3OH, ΔfH = -239 kJ mol-1

 

 

Question 6.15

Calculate the enthalpy change for the process

CCl4 (g) --> C (g) + 4 Cl(g)

And calculate bond enthalpy of C –Cl in CCl4 (g).

ΔvapH (CCl4) = 30.5 kJ mol-1

ΔfH (CCl4) = -135.5 kJ mol-1

ΔaH (C) = 715.0 kJ mol-1, where ΔaH is enthalpy of atomization

ΔaH (Cl2) = 242 kJ mol-1

Answer.

CCl4 (l) --> CCl4 (g), ΔvapH = 30.5 kJ mol-1 –(1)

C (s) + 2Cl2 (g) --> CCl4 (l), ΔfH = - 135.5 kJ mol-1 –(ii)

C (s) --> C (g), ΔaH = 715 kJ mol-1 –(iii)

Cl2 (g) -->2 Cl (g), ΔaH =242 kJ mol-1 –(iv)

Now, we aim at equation

CCl4 (g) --> C (g) + 4Cl (g)

Equation (iii) + 2 x Equation (iv) – Equation (i)

ΔH = 715.0 + 2 (242) – 30.5 – (-135.5) kJ mol-1 = 1304 kJ mol-1

Bond enthalpy  of C-Cl  = (1304/4) = 326 kJ mol-1.

 

 

Question 6.16

For an isolated system, ΔU = 0, what will be ΔS?

Answer.

ΔU is positive, ΔU > 0

ΔU = 0 then ΔS will be positive due to which reaction will be spontaneous.

 

 

Question 6.17

For a reaction at 298 K,

2A + B --> C

ΔH = 400kJ mol-1 and ΔS = 0.2 kJ K-1 mol-1.

At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range.

Answer.

According to Gibbs Helmholtz equation,

ΔG = (ΔH – T ΔS)

For ΔG = 0, ΔH = T ΔS

T = (ΔH/ ΔS)

T = (400 kJ mol-1)/(0.2 kJ mol-1 K-1) = 2000 K

 

 

Question 6.18

For the reaction,

2 Cl(g) --> Cl2 (g), what are the signs of ΔH and ΔS?

Answer.

ΔH will be negative because energy is released in bond formation.

ΔS will be negative because entropy decreases when atoms combine to form molecules.

 

Question 6.19

For the reaction

2 A (g) + 2 B (g) --> 2 D (g)

ΔU = -10.5 kJ and ΔS = -44.1 JK-1.

Calculate ΔG for the reaction, and predict whether the reaction may occur spontaneously.

Answer.

2 A (g) + 2 B (g) --> 2 D (g)

Δng = 2 – 3 = -1 mole

Put the value of ΔU in the expression of ΔH

ΔH = ΔU + Δng RT

ΔH = (-10.5 kJ  - (-1) (8.314 x 10-3 kJ mol-1 K-1) (298 K)

= -10.5 kJ – 2.48 kJ = -12.98 kJ

 

Putting the values of ΔS and ΔH in the expression of ΔG

ΔG = ΔH – T ΔS

-12.98 – (298 K) (-44.1 J K-1)

- 12.98 kJ + 13.14 kJ = 0.16 kJ

ΔG is positive, so reaction will not occur spontaneously.

 

 

Question 6.20

The equilibrium constant for a reaction is 10. What will be the value of ΔG?

R = 8.314 JK-1, T = 300 K

Answer.

ΔG = -2.303 RT log eq

= (2.303) x (8.314 x 10-3 kJ mol-1 K-1) x (300 K) log 10

= -5744.14 J mol-1

 

 

Question 6.21

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

(½) N2 (g) + ½ O2 (g) --> NO (g), Δf H = 90 kJ mol-1

NO (g) + (½) O2 (g) --> NO (g), Δf H = -74 kJ mol-1

Answer.

Δf H is positive for NO, it is unstable in nature.

Δf H is negative for NO2 , it is stable in nature.

 

 

Question 6.22

Calculate the entropy change in surroundings when 1.00 mol of H2O (l) is formed under standard conditions.

Δf H = -286 kJ mol-1.

Answer.

Q = - 286 kJ mol-1 (given so that amount of heat is evolved during formation of 1 mole of water)

Same heat will be absorbed by surrounding. Q = + 286 kJ mol-1

ΔS  = (Q/7) = (286 kJ mol-1/298 K)

ΔS = 959.73 J mol-1 K-1

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