# Thermodynamics

## NCERT Solution

### Part 1

Question 1: A thermodynamic state function is a quantity

1. Used to determine heat changes
2. Whose value is independent of path
3. Used to determine pressure volume work
4. Whose value depends on temperature only

Answer: (b) Whose value is independent of path

Question 2: For the process to occur under adiabatic conditions, the correct condition is

1. ΔT = 0
2. Δp = 0
3. q = 0
4. w = 0

Question 3: The enthalpies of all elements in their standard states are

1. Unity
2. Zero
3. < zero
4. Different for each element

Question 4: ΔUΘ of combustion of methane is – X kJ mol-1. The value of ΔHΘ is

1. = ΔUΘ
2. > ΔUΘ
3. < ΔUΘ
4. = 0

Answer: Balanced chemical equation for combustion of methane is as follows:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

In this reaction, there are 3 moles of gaseous reactants and 1 mole of gaseous products

So, Δng = 1 – 3 = - 2

We know, ΔH = ΔU + ΔngRT

Substituting the values, we have:

ΔH = ΔU – 2RT

Or, ΔU = ΔH + 2RT

This equation means that ΔH < ΔU

So correct answer is option (c)

Question 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. Enthalpy of formation of CH4(g) will be

1. -74.8 kJ mol-1
2. -52.27 kJ mol-1
3. +74.8 kJ mol-1
4. +52.26 kJ mol-1

Answer: Balanced equation for combustion of methane, graphite and dihydrogen are given below in respective order

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

C(g) + O2(g) → CO2(g)

H2 + 1/2O2(g) → H2O(l)

We can get the equation for formation of methane if add above equations as follows:

Equation (2) + 2 × equation (3) - equation (1)

C + O2 + 2(H2 + 1/2O2) – (CH4 + 2O2)

= C + O2 + 2H2 + O2 - CH4 - 2O2

= C + 2H2 - CH4

CO2 + 2H2O – CO2 - 2H2 = 0

So, C + 2H2 - CH4 = 0

Or, C + 2H2 = CH4

Now, enthalpy of formation of methane can be calculated as follows:

Equation (2) + 2 × equation (3) - equation (1)

= -393.5 – 2 × 285.8 + 890.3 = -393.5 – 571.6 + 890.3 = -74.8 kJ mol-1

So, correct answer is option (a)

Question 6: A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be

1. Possible at high temperature
2. Possible only at low temperature
3. Not possible at any temperature
4. Possible at any temperature

Answer: (d) Possible at any temperature

Question 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: ΔU = q + w

= 701 – 394 (work is negative because it is done by the system)

= 307 J

Question 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 the enthalpy change for the reaction at 298 K.

NH2CN(s) + 3/2O2(g) → CO2(g) + H2O(l)

Answer: Given, ΔU = -742.7 kJ mol-1, T = 298 K.

R = 8.314 × 10-3 J K-1mol-1

Δng = 2-3/2=1/2 mol

ΔH = ΔU + ΔngRT

= -742.7 + 1/2xx8.314xx10^(-3)xx298

=-742.70.004157xx298=-742.7+1.24=741.46 kJ

Question 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: Number of moles of Al (m) = (60)/(27)=2.22 mol

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

ΔT = 55 – 35 = 20°C = 20 K

Heat evolved q = C × m × T

= 24 × 2.22 × 20 = 1065.6 J = 1.067 kJ

Question 10: 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

Answer: This change can be represented by following diagram:

According to Hess's Law:

ΔH = ΔH1 + ΔH2 + ΔH3

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

ΔH2 = -6.03 kJ mol-1 = -6030 J mol-1

(sign changed to negative due to solidification)

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

Now, ΔH = 753 – 6030 – 368 = -5645 J = - 5.654 kJ mol-1

Question 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: Balanced equation for combustion of carbon is as follows:

C(s) + O2 → CO2(g)

Given: ΔH = - 393.5 kJ

Molar mass of carbon dioxide = 44 g

Since combustion of 44 g of CO2 gives 393.5 kJ

Hence, combustion of 356.2 g of CO2 will give

(393.5)/(44)xx35.2=314.8 kJ