BITSAT Chemistry Thermodynamics MCQs

Question: For the reaction  Which one of the statement is correct at constant T and P ?

a) ΔH=ΔE

b) ΔH< ΔE

c) ΔH>ΔE

d) ΔH is independent of physical state of the reactants

Answer: ΔH< ΔE

Question: For a given reaction, ΔH = 35.5 kJ mol^-1 and ΔS = 83.6 JK^-1 mol^-1 1. The reaction is spontaneous at : (Assume that ΔH and ΔS do not vary with tempearature)

a) T > 425 K

b) All temperatures

c) T > 298 K

d) T < 425 K

Answer: T > 425 K

Question: For a particular reversible reaction at temperature T, ΔH and ΔS were found to be both +ve. If T_e is the temperature at equilibrium, the reaction would be spontaneous when

a) T_e > T

b) T > T_e

c) T_e is 5 times T

d) T = T_e

Answer: T > T_e

Question: Given

Based on data provided, the value of electron gain enthalpy of fluorine would be :

a) – 300 kJ mol^–1

b) – 350 kJ mol^–1

c) – 328 kJ mol^–1

d) – 228 kJ mol^–1

Answer: – 328 kJ mol^–1

Question: Which law of the thermodynamics helps in calculating the absolute entropies of various substances at different temperatures?

a) First law

b) Second law

c) Third law

d) Zeroth law

Answer: Third law

Question: A spontaneous reaction is impossible if

a) Both ΔH and ΔS are negative

b) Both ΔH and ΔS are positive

c) ΔH is negative and ΔS is positive

d) ΔH is positive and ΔS is negative

Answer: ΔH is positive and ΔS is negative

Question: If the enthalpy of vaporization of water is 186.5 kJmol^–1, the entropy if its vaporization will be :

a) 0.5 JK^–1mol^–1

b) 1.0 JK^–1mol^–1

c) 1.5 JK^–1mol^–1

d) 2.0 JK^–1mol^–1

Answer: 0.5 JK^–1mol^–1

Question: The heats of neutralisation of CH₃COOH, HCOOH, HCN and H₂S are – 13.2, – 13.4, – 2.9 and – 3.8 kCal per equivalent respectively. Arrange the acids in increasing order of acidic strength.

a) HCOOH > CH₃COOH > H₂S > HCN

b) CH₃COOH > HCOOH > H₂S > HCN

c) H₂S > HCOOH > CH₃COOH > HCN

d) HCOOH > H₂S > CH₃COOH > HCN

Answer: HCOOH > CH₃COOH > H₂S > HCN

Question: For determining the spontaneity of a process which of the following is considered ?

a) ΔS system

b) ΔS surroundings

c) ΔS system + ΔS surroundings

d) ΔS system – ΔS surroundings

Answer: ΔS system + ΔS surroundings

Question: The enthalpy of combustion of 2 moles of benzene at 27°C differs from the value determined in bomb calorimeter by

a) – 2.494 kJ

b) 2.494 kJ

c) – 7.483 kJ

d) 7.483 kJ

Answer: – 7.483 kJ

Question: In which of the following reactions, standard entropy change (ΔS°) is positive and standard Gibb’s energy change (ΔG°) decreases sharply with increasing temperature ?

a)

b)

c)

d)

Answer:

Question: Bond enthalpies of H₂ , X₂ and HX are in the ratio 2 : 1 : 2. If enthalpy of formation of HX is – 50 kJ mol^–1, the bond enthalpy of X₂ is

a) 100 kJ mol^–1 

b) 300 kJ mol^–1

c) 200 kJ mol^–1

d) 400 kJ mol^–1

Answer: 100 kJ mol^–1 

Question: What is the free energy change, ΔG , when 1.0 mole of water at 100º C and 1 atm pressure is converted into steam at 100°C and 1 atm. pressure?

a) 540 cal

b) –9800 cal

c) 9800 cal

d) 0 cal

Answer: 0 cal

Question: For the reaction :  Which one of the following is correct regarding ΔH

a) ΔH = ΔE + 2RT

b) ΔH = ΔE – 2RT

c) ΔH = ΔE + RT

d) ΔH = ΔE – RT

Answer: ΔH = ΔE – 2RT

Question: One mole of an ideal gas at 300 K is expanded isothermally from an initial volume of 1L to 10 L. The ΔE for this process is (R = 2 cal mol^–1 K^–1)

a) 163.7 cal

b) zero

c) 1381.1 cal

d) 9 lit atom

Answer: zero

Question: At 25°C and 1 bar which of the following has a non-zero ΔH^°_f ?

a) Br₂(l)

b) C (graphite)

c) O₃(g)

d) I₂(s)

Answer: O₃(g)

Question: For which of the process, ΔS is negative?

a) H₂ (g)→ 2H(g)

b) N₂ (g )(1atm)→ N₂ (g)(8atm)

c) 2SO₃ (g)→ 2SO₂ (g) +O₂ (g)

d) C_(diamond) → C_(graphite)

Answer: N₂ (g )(1atm)→ N₂ (g)(8atm)

Question: K_a for CH₃COOH at 25°C is 1.754 × 10^–5 . At50°C K_a is 1.633 × 10^–5. What are ΔH° and ΔS° for the ionisation of CH₃COOH ?

a) – 1.55 kJmol^-1, – 96.44 Jmol ^-1K^-1

b) – 2.55 kJmol^-1, – 106.44 Jmol ^-1K^-1

c) 1.55 kJmol^-1, 96.44 Jmol ^-1K^-1

d) – 1.55 kJmol^-1, 96.44 Jmol ^-1K^-1

Answer: – 1.55 kJmol^-1, – 96.44 Jmol ^-1K^-1

Question: Temperature of 5 moles of a gas is decreased by 2K at constant pressure. Indicate the correct statement

a) Work done by gas is = 5 R

b) Work done over the gas is = 10 R

c) Work done by the gas = 10 R

d) Work done = 0

Answer: Work done over the gas is = 10 R

Question: Compounds with high heat of formation are less stable because

a) It is difficult to synthesize them

b) Energy rich state leads to instability

c) High temperature is required to synthesize them

d) Molecules of such compunds are distorted

Answer: Energy rich state leads to instability

Question: The energy that opposes dissolution of a solvent is

a) Hydration energy

b) Lattice energy

c) Internal energy

d) Bond energy

Answer: Lattice energy

Question: The energy that opposes dissolution of a solvent is

a) Hydration energy

b) Lattice energy

c) Internal energy

d) Bond energy

Answer: Lattice energy

Question: In the reaction  ΔH = 40 kJ; ΔH represents :

a) Heat of formation

b) Heat of combustion

c) Heat of neutralisation

d) Heat of reaction

Answer: Heat of reaction

Question: The energy that opposes dissolution of a solvent is

a) Hydration energy

b) Lattice energy

c) Internal energy

d) Bond energy

Answer: Lattice energy

Question: Which law of the thermodynamics helps in calculating the absolute entropies of various substances at different temperatures?

a) First law

b) Second law

c) Third law

d) Zeroth law

Answer: Third law