Question 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

Accepted Answer

A thermodynamic state function is a quantity whose value is independent of a path. Functions like p, V, T etc. depend only on the state of a system and not on the path. Hence, alternative (ii) is correct.

Question 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

Accepted Answer

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.

Question 3

The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element

Accepted Answer

The enthalpy of all elements in their standard state is zero. Therefore, alternative (ii) is correct.

Question 4

ΔU⊖ of combustion of methane is −X kJ mol−1. The value of ΔH⊖ is
(i) =ΔU⊖
(ii) >ΔU⊖
(iii) <=ΔU⊖
(iv) = 0

Accepted Answer

Since ΔHθ = ΔUθ + ΔngRT and ΔUθ = −X kmol−1, then ΔHθ = (−X) + ΔngRT.  For the combustion of methane, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), Δng = 1 - 3 = -2.  Thus, ΔHθ = ΔUθ - 2RT. This implies that ΔHθ < ΔUθ.  Therefore, alternative (iii) is correct.

Question 5

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3kJmol−1 −393.5kJmol−1, and −285.8kJmol−1 respectively. Enthalpy of formation of CH4 will be
(i) −74.8kJmol−1
(i

Accepted Answer

According to the question, the given reactions are: (i) CH₄(g)+2O₂(g)⟶CO₂(g)+2H₂O(l) ΔH=−890.3kJmol−1 (ii) C(s)+O₂(g)⟶CO₂(g) ΔH=−393.5kJmol−1 (iii) H₂(g)+1/2O₂(g)⟶H₂O(l) ΔH=−285.8kJmol−1 The desired equation is the one that represents the formation of CH₄(g): C(s) + 2H₂(g) → CH₄(g). By combining the given equations (equation (ii) + 2 × equation (iii) - equation (i)), the enthalpy of formation of CH₄(g) is calculated as [-393.5 + 2(-285.8) - (-890.3)] kJmol⁻¹ = -74.8 kJmol⁻¹. Hence, alternative (

Question 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

Accepted Answer

For a reaction to be spontaneous, ΔG should be negative. The formula is ΔG = ΔH – TΔS. According to the question, for the given reaction, ΔS = positive and ΔH = negative (since heat 'q' is evolved). This means ΔG will always be negative. Therefore, the reaction is spontaneous at any temperature. Hence, alternative (iv) is correct.

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?

Accepted Answer

According to the first law of thermodynamics, ΔU = q + W. Given: q = +701 J (since heat is absorbed) W = –394 J (since work is done by the system) Substituting the values: ΔU = 701 J + (–394 J) = 307 J. Hence, the change in internal energy for the given process is 307 J.

Question 8

The reaction of cyanamide, NH₂CN(s) with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7kJmol−1 at 298 K. Calculate enthalpy change for the reaction at 298 K.
NH₂CN(g)

Accepted Answer

Enthalpy change for a reaction (ΔH) is given by ΔH = ΔU + ΔngRT. For the given reaction, Δng = ∑ng(products) – ∑ng(reactants) = (1 + 1) – (1 + 1.5) = 2 – 2.5 = -0.5 moles. Given: ΔU = –742.7 kJ mol–1, T = 298 K, R = 8.314 × 10–3 kJ mol–1 K–1. Substituting the values: ΔH = (–742.7 kJ mol–1) + (-0.5 mol) (298 K) (8.314 × 10–3 kJ mol–1 K–1) = –742.7 - 1.239 = –743.9 kJ mol–1.

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 Jmol−1K−1.

Accepted Answer

From the expression of heat (q), q = n ⋅ c ⋅ ΔT. Where c = molar heat capacity, n = moles of substance, ΔT = change in temperature. Given: mass of Al = 60.0 g, molar mass of Al = 27 g/mol, c = 24 Jmol–1K–1, ΔT = 55°C - 35°C = 20°C = 20 K. Moles of Al (n) = 60.0g / 27g/mol. Substituting the values: q = (60/27 mol) × (24 Jmol−1K−1) × (20 K) = 1066.67 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=6.03kJmol−1 at 0∘C
Cp[H₂O(l)]=75.3Jmol−1K−1
Cρ[H₂O(s)]=36.8Jmol−1K−1

Accepted Answer

Total enthalpy change involved is the sum of three changes: (a) Energy change for 1 mol of water at 10°C to 1 mol of water at 0°C. (b) Energy change for 1 mol of water at 0°C to 1 mol of ice at 0°C (freezing). (c) Energy change for 1 mol of ice at 0°C to 1 mol of ice at –10°C. The formula for total enthalpy change (ΔH) is: ΔH = Cp[H₂O(l)]ΔT₁ + ΔHfreezing + Cρ[H₂O(s)]ΔT₂. Note: ΔHfreezing = -ΔfusH = -6.03 kJmol⁻¹ = -6030 Jmol⁻¹. Substituting the values: ΔH = (75.3 Jmol⁻¹K⁻¹)(0−10)K + (−6030 Jmol⁻

Thermodynamics