# Thermodynamics

## NCERT Solution

### Part 1

Question 1: 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:** (b) Whose value is independent of path

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

- ΔT = 0
- Δp = 0
- q = 0
- w = 0

**Answer:** (c) q = 0

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

- Unity
- Zero
- < zero
- Different for each element

**Answer:** (b) zero

Question 4: ΔU^{Θ} of combustion of methane is – X kJ mol^{-1}. The value of ΔH^{Θ} is

- = ΔU
^{Θ} - > ΔU
^{Θ} - < ΔU
^{Θ} - = 0

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

CH_{4}(g) + 2O_{2}(g) → CO_{2}(g) + 2H_{2}O(l)

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

So, Δn_{g} = 1 – 3 = - 2

We know, ΔH = ΔU + Δn_{g}RT

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 CH_{4}(g) will be

- -74.8 kJ mol
^{-1} - -52.27 kJ mol
^{-1} - +74.8 kJ mol
^{-1} - +52.26 kJ mol
^{-1}

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

CH_{4}(g) + 2O_{2}(g) → CO_{2}(g) + 2H_{2}O(l)

C(g) + O_{2}(g) → CO_{2}(g)

H_{2} + `1/2`O_{2}(g) → H_{2}O(l)

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

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

Addition of LHS of equations:

C + O_{2} + 2(H_{2} + `1/2`O_{2}) – (CH_{4} + 2O_{2})

= C + O_{2} + 2H_{2} + O_{2} - CH_{4} - 2O_{2}

= C + 2H_{2} - CH_{4}

Addition of RHS of equations:

CO_{2} + 2H_{2}O – CO_{2} - 2H_{2} = 0

So, C + 2H_{2} - CH_{4} = 0

Or, C + 2H_{2} = CH_{4}

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

- Possible at high temperature
- Possible only at low temperature
- Not possible at any temperature
- 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, NH_{2}CN (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.

NH_{2}CN(s) + `3/2`O_{2}(g) → CO_{2}(g) + H_{2}O(l)

**Answer:** Given, ΔU = -742.7 kJ mol^{-1}, T = 298 K.

R = 8.314 × 10^{-3} J K^{-1}mol^{-1}

Δn_{g} = `2-3/2=1/2` mol

ΔH = ΔU + Δn_{g}RT

`= -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

Δ_{fus}H = 6.03 kJ mol^{-1} at 0°C.

C_{p}[H_{2}O(l)] = 75.3 J mol^{-1} K^{-1}

C_{p}[H_{2}O(s)] = 36.8 J mol^{-1} K^{-1}

**Answer:** This change can be represented by following diagram:

According to Hess's Law:

ΔH = ΔH_{1} + ΔH_{2} + ΔH_{3}

Now, ΔH_{1} = 75.3 J mol^{-1} K^{-1}(10 K) = 753 J mol^{-1}

ΔH_{2} = -6.03 kJ mol^{-1} = -6030 J mol^{-1}

(sign changed to negative due to solidification)

ΔH_{3} = 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 CO_{2} is -393.5 kJ mol^{-1}. Calculate the heat released upon formation of 35.2 g of CO_{2} from carbon and dioxygen gas.

**Answer:** Balanced equation for combustion of carbon is as follows:

C(s) + O_{2} → CO_{2}(g)

Given: ΔH = - 393.5 kJ

Molar mass of carbon dioxide = 44 g

Since combustion of 44 g of CO_{2} gives 393.5 kJ

Hence, combustion of 356.2 g of CO_{2} will give

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