# CBSE Class 11 Chemistry Thermodynamics MCQ with Answers PDF

CBSE MCQ Questions for Class 11 Chemistry Chapter 6 Thermodynamics with Answers Pdf free download. MCQ Questions for Class 11 Chemistry with Answers were prepared based on the latest exam pattern. This MCQ will help you score good marks in the final exam. Thermodynamics Class 11 Chemistry MCQs are prepared for a better understanding of the concept. Thermodynamics Class 11 Chemistry MCQ is prepared by Experts of CBSE.

# Class 11 Chemistry Chapter 6 Thermodynamics MCQs

Check the multiple-choice questions for the 11th Class Chemistry Chapter 6 Thermodynamics. Select one answer out of 4 options.

## Thermodynamics Class 11 MCQs Questions with Answers

Amount of heat required to change 1 g ice at 0°C to 1 g steam at 100°C is
(a) 616 cal
(b) 12 R cal
(c) 717 cal
(d) None of these

Which thermodynamic function accounts automatically for enthalpy and entropy both?
(a) Helmholtz Free Energy (A)
(b) Internal Energy (E)
(c) Work Function
(d) Gibbs Free Energy

One gram of sample of NH4NO3 is decomposed in a bomb calorimeter. The temperature of the calorimeter increases by 6.12 K. The heat capacity of the system is 1.23 kj/deg. What is the molar heat of decomposition of NH4NO3?
(a) -7.53 kj/mol
(b) -398.1 kj mol-1
(c) -16.1 kj/mol
(d) -602 kj/mol

Entropy of the universe is
(a) Continuously Increasing
(b) Continuously Decreasing
(c) Zero
(d) Constant

The amount of the heat released when 20 ml 0.5 M NaOH is mixed with 100 ml 0.1 M HCl is x kJ. The heat of neutralization is
(a) -100 × kJ/mol
(b) -50 × kJ/mol
(c) 100 × KJ/mol
(d) 50 × kJ/mol

C(diamond) → C(graphite) ∆H = -ve. This shows that
(a) Graphite is more stable than diamond
(b) Diamond is more stable than graphite
(c) Both are equally stable
(d) Stability cannot be predicted.

Answer: (a) Graphite is more stable than diamond

AHr of graphite is 0.23 kj/mol and ∆Hf for diamond is 1.896 kj mol-1, ∆Htransition from graphite to diamond is
(a) 1.66 kj/mol
(b) 2.1 kj/mol
(c) 2.33 kj/mol
(d) 1.5 kj/mol

The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283.0 kJ mol-1 respectively. The enthalpy of formation of carbon monoxide is:
(a) -676 kJ
(b) 110.5 kJ
(c) -110.5 kJ
(d) 676.5 kJ

One mole of which of the following has the highest entropy?
(a) Liquid Nitrogen
(b) Hydrogen Gas
(c) Mercury
(d) Diamond

The bond energies of C-C, C=C; H-H and C-H linkages are 350, 600, 400 and 410 kj per mole respectively. The heat of hydrogenation of ethylene is
(a) -170 kj mol-1
(b) -260 kj mol-1
(c) 400 kj mol-1
(d) -450 kj mol-1

For an ideal gas, CV and CP are related as :
(a) CV – CP = R
(b) CV + CP = R
(c) CP – Cv = RT
(d) CP – Cv = R

Answer: (d) CP – Cv = R

The species which by definition has ZERO standard molar enthalpy of formation at 298 K is
(a) Br2(g)
(b) Cl2(g)
(c) H2O(g)
(d) CH4(g)

What is the entropy change (in JK-1 mol-1) when 1 mole of ice is converted into water at 0℃? (The enthalpy change for the conversion of ice to liquid water is 6.0 kJ mol-1 at 0℃)
(a) 20.13
(b) 2.013
(c) 2.198
(d) 21.98

Enthalpy of CH4 + 1/2 O2 → CH3OH is negative.
If enthalpy of combustion of CH4 and CH3OH are x and y respectively then which reaction is correct?
(a) x > y
(b) x < y
(c) x = y
(d) x ≥ y

Which of the following is true for the reaction? H2O (l) ↔ H2O (g) at 100° C and 1 atm pressure
(a) ∆S = 0
(b) ∆H = T ∆S
(c) ∆H = ∆U
(d) ∆H = 0

Standard enthalpy of vapourisation ΔHvap for water at 100°C is 40.66 kJmol-1. The internal energy of vapourisation of water at 100°C (in kJmol-1) is (Assume water vapour to behave like an ideal gas)
(a) 43.76
(b) 40.66
(c) 37.56
(d) -43.76

Heat engine is a device by which a system is made to undergo a …X… process that result in conversion of …Y… into work. Here, X and Y refer to
(a) cyclic and heat
(b) None of these
(c) isothermal and heat
(d) cyclic and work

Molar heat capacity of water in equilibrium with ice at constant pressure is
(a) zero
(b) infinity
(c) 40.45 kj K-1 mol-1
(d) 75.48 JK-1 mol-1

Which one of the following statement is false?
(a) Work is a state function
(b) Temperature is a state function
(c) Change in the state is completely defined when the initial final states are specified
(d) Work appears at the boundary of the system

Answer: (a) Work is a state function

Calculate the heat required to make 6.4 Kg CaC2 from CaO(s) and C(s) from the reaction: CaO(s) + 3 C(s) → CaC2(s) + CO (g) given that ∆f Ho (CaC2) = -14.2 kcal. ∆f H° (CO) = -26.4 kcal.
(a) 5624 kcal
(b) 1.11 × 104 kcal
(c) 86.24 × 10³
(d) 1100 kcal

Answer: (b) 1.11 × 104 kcal

A system absorb 10 kJ of heat at constant volume and its temperature rises from 270 C to 370 C. The value of ∆ U is
(a) 100 kJ
(b) 10 kJ
(c) 0 kJ
(d) 1 kJ

Hess’s law is an application of
(a) 1st law of Thermodynamics
(b) 2nd law of Thermodynamics
(c) Entropy change
(d) ∆H = ∆U + P∆V

Answer: (a) 1st law of Thermodynamics

The bond energy (in kcal mol-1) of a C-C single bond is approximately
(a) 1
(b) 10
(c) 83-85
(d) 1000

Which of the following has the highest entropy?
(a) Mercury
(b) Hydrogen
(c) Water
(d) Graphite

One mole of a hon-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K) → (4.0 atm, 5.0 L, 245 K) with a change in internal energy, ∆U = 30.0 L atm. The change in enthalpy (∆H) of the process in L atm is
(a) 44.0
(b) 42.3
(c) 44.0
(d) not defined because pressure is not constant.

In what proportional 1 M NaOH and 0.5 MH2SO4 are mixed respectively so as to release maximum amount of energy and to form 100 ml of solution?
(a) 33 and 67
(b) 67 and 33
(c) 40 and 60
(d) 50 and 50

The temperature of source and sink of a heat engine are 127ºC and 27ºC respectively. An inventor claims its efficiency to be 26%, then
(a) it is impossible
(b) it is possible with high probability
(c) it is possible with low probability
(d) data are insufficient

If the co-efficient of performance of a refrigerator is 5 and operates at the room temperature 27°C, the temperature inside the refrigerator is
(a) 250 K
(b) 230 K
(c) 260 K
(d) 240 K

Which of the following reaction defines ∆H0f?
(a) C(Diamond) + O2(g) → CO2(g)
(b) 1/2 H2(g) + 1/2 F2(g) → HF(g)
(c) N2(g) + 3H2(g) → 2NH3(g)
(d) CO(g) + 1/2 O2(g) → CO2(g)

Answer: (b) 1/2 H2(g) + 1/2 F2(g) → HF(g)

Air conditioner is based on the principle of
(a) refrigerator
(b) first law of thermodynamics
(c) None of these
(d) Carnot cycle