# NCERT Solutions for Class 11 Chemistry Chemistry Part-1 Chapter 6

## Thermodynamics Class 11

### Exercise : Solutions of Questions on Page Number : 182

Q1 :

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.

A thermodynamic state function is a quantity whose value is independent of a path.

Functions like p, V, Tetc. depend only on the state of a system and not on the path.

Hence, alternative (ii) is correct.

Q2 :

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

(i) ΔT = 0

(ii) Δp = 0

(iii) q = 0

(iv) w= 0

A system is said to be under adiabatic conditions if there is no exchange of heat between the system and its surroundings. Hence, under adiabatic conditions, q = 0.

Therefore, alternative (iii) is correct.

Q3 :

The enthalpies of all elements in their standard states are:

(i) unity

(ii) zero

(iii) < 0

(iv) different for each element

Q4 :

ΔUÃŽÂ¸of combustion of methane is - XkJ mol-1. The value of ΔHÃŽÂ¸is

(i) = ΔUÃŽÂ¸

(ii) > ΔUÃŽÂ¸

(iii) < ΔUÃŽÂ¸

(iv) = 0

Q5 :

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 kJ mol-1             (ii) -52.27 kJ mol-1

(iii) +74.8 kJ mol-1           (iv) +52.26 kJ mol-1.

Q6 :

A reaction, A + B → C + D + qis 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

Q7 :

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

Q8 :

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-1at 298 K. Calculate enthalpy change for the reaction at 298 K.

Q9 :

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

Q10 :

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

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

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

Q11 :

Enthalpy of combustion of carbon to CO2is -393.5 kJ mol-1. Calculate the heat released upon formation of 35.2 g of CO2from carbon and dioxygen gas.

Q12 :

Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are â€“110 kJ molâ€“1, â€“ 393 kJ molâ€“1, 81 kJ molâ€“1and 9.7 kJ molâ€“1respectively. Find the value of ΔrH for the reaction:

N2O4(g)+ 3CO(g) N2O(g)+ 3CO2(g)

Q13 :

Given

; ΔrHθ= â€“92.4 kJ molâ€“1

What is the standard enthalpy of formation of NH3gas?

Q14 :

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

CH3OH(l) + O2(g) CO2(g) + 2H2O(l) ; ΔrHθ = â€“726 kJ molâ€“1

C(g) + O2(g) CO2(g) ; ΔcHθ = â€“393 kJ molâ€“1

H2(g) +

Q15 :

Calculate the enthalpy change for the process

CCl4(g) → C(g) + 4Cl(g)

and calculate bond enthalpy of C-Cl in CCl4(g).

ΔvapHÃŽÂ¸ (CCl4) = 30.5 kJ mol-1.

ΔfHÃŽÂ¸ (CCl4) = -135.5 kJ mol-1.

ΔaHÃŽÂ¸ (C) = 715.0 kJ mol-1, where ΔaHÃŽÂ¸ is enthalpy of atomisation

ΔaHÃŽÂ¸ (Cl2) = 242 kJ mol-1

Q16 :

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

Q17 :

For the reaction at 298 K,

2A + B → C

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

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

Q18 :

For the reaction,

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

Q19 :

For the reaction

2A(g) + B(g) → 2D(g)

ΔUÃŽÂ¸ = -10.5 kJ and ΔSÃŽÂ¸= -44.1 JK-1.

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

Q20 :

The equilibrium constant for a reaction is 10. What will be the valueof ΔGÃŽÂ¸? R = 8.314 JK-1mol-1, T= 300 K.

Q21 :

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

N2(g) + O2(g) Ã¢â€ ’ NO(g); ΔrHθ= 90 kJ molâ€“1

NO(g)+O2(g) Ã¢â€ ’ NO2(g) : ΔrHθ= â€“74 kJ molâ€“1