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MCQ Questions Class 11 Chemistry Thermodynamics with Answers - Set - 2
Question 1:
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.
Correct Answer – (D)
Question 2 :
Which of the following salts will have maximum cooling effect when 0.5 mole of the salt is dissolved in same amount of water. Integral heat of solution at 298 K is given for each salt?
(a) KNO3 (∆H = 35.4 kJ mol-1)
(b) NaCl (∆H = 5.35 kJ mol-1)
(c) HBr (∆H = -83.3 kJ mol-1)
(d) KOH ( ∆H = -55.6 kJ mol-1)
Correct Answer – (A)
More the heat absorbed, more will be the cooling effect. Hence, more the positive value of ∆ H, more the cooling effect
Question 3 :
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
Correct Answer – (B)
At constant volume w = 0
Therefore, ∆ U = q = 10 kJ
Question 4 :
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
Correct Answer – (A)
Normality of NaOH = Molarity × acidity
= 0.5 × 1 = 0.5 N
Total heat q produced = x kJ
Heat of neutralisation
= [(q)/ (Volume of acid or base)] ×1000× (1/normality of acid or base)
= (x/20) × 1000 × (1/0.5)
= 100 x kJmol-1
Since heat is liberated, heat of neutralisation = −100 x kJmol-1
Question 5 :
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
Correct Answer – (D)
Gibbs free energy combines the effect of both enthalpy and entropy. The change in free energy (ΔG) is equal to the sum of the change of enthalpy (∆H) minus the product of the temperature and the change of entropy (∆S) of the system.
∆G = ∆H – T∆S
ΔG predicts the direction in which a chemical reaction will go under two conditions: (1) constant temperature and (2) constant pressure.
If ΔG is positive, then the reaction is not spontaneous (it requires the input of external energy to occur) and if it is negative, then it is spontaneous (occurs without the input of any external energy)
MCQ Questions Class 11 Chemistry Thermodynamics with Answers
Question 6 :
Standard enthalpy of vapourisation ΔHvap for water at 100oC 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
Correct Answer – (C)
For gaseous reactants and products, we have a relation between standard enthalpy of vaporization (ΔHvap) and standard internal energy (ΔE) as-
ΔHvap = ΔE+ Δng RT whereas,
Δng = n2 − n1, i.e., difference between no. of moles of reactant and product.
For vaporization of water,
H2O (l) → H2O(g)
Therefore, Δng = 1 − 0 = 1
Therefore, ΔHvap = ΔE+ Δng RT
⇒ ΔE = ΔHvap − Δng
RT = 40.63 − (1×8.314×10-3 × 373) = 37.53 KJ/mol
Hence the value ΔE for this process will be 37.53KJ/mol
Question 7 :
The temperature of the system decreases in an ______.
(a) Adiabatic Compression
(b) Isothermal Expansion
(c) Isothermal Compression
(d) Adiabatic Expansion
Correct Answer – (D)
In adiabatic process heat is neither added nor removed from system. So the work done by the system (expansion) in adiabatic process will result in decrease of internal energy of that system (from first law).
As internal energy is directly proportional to the change in temperature there will be temperature drop in an adiabatic process
Question 8 :
Based on the first law of thermodynamics, which one of the following is correct?
(a) For an isothermal process, q = +w
(b) For an isochoric process, ΔU = -q
(c) For an adiabatic process, ΔU = -w
(d) For a cyclic process, q = -w
Correct Answer – (D)
(1) ΔU = q + w. For an isochoric process, w = −PΔV = 0. Hence, ΔU = qv
(2) For an adiabatic process, q = 0. Hence, ΔU = w
(3 ) For an isothermal process, ΔU = 0 Hence, q = −w
(4) For a cyclic process , ΔU = 0. Hence, q = −w
Question 9 :
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
Correct Answer – (A)
Question 10 :
The bond energy (in kcal mol-1) of a C-C single bond is approximately
(a) 1
(b) 10
(c) 83-85
(d) 1000
Correct Answer – (C)
Explanation:
C–C bond 83–85 kcal/mol
It is the energy required to break the bond .It is defined as the standard enthalpy change when a bond is cleaved by homolysis, with reactants and products of the homolysis reaction at 0 K (absolute zero)
- NCERT Solutions Class 11 Chemistry Chapter 1 : Some Basic Concepts of Chemistry
- NCERT Solutions Class 11 Chemistry Chapter 2 : Structure Of The Atom
- NCERT Solutions Class 11 Chemistry Chapter 3 : Classification of Elements and Periodicity in Properties
- NCERT Solutions Class 11 Chemistry Chapter 4 : Chemical Bonding and Molecular Structure
- NCERT Solutions Class 11 Chemistry Chapter 5 : States of Matter
- NCERT Solutions Class 11 Chemistry Chapter 6 : Thermodynamics
- NCERT Solutions Class 11 Chemistry Chapter 7 : Equilibrium
- NCERT Solutions Class 11 Chemistry Chapter 8 : Redox Reactions
- NCERT Solutions Class 11 Chemistry Chapter 9 : Hydrogen
- NCERT Solutions Class 11 Chemistry Chapter 10 : The s-Block Elements
- NCERT Solutions Class 11 Chemistry Chapter 11 : The p-Block Elements
- NCERT Solutions Class 11 Chemistry Chapter 12 : Organic Chemistry: Some Basic Principles and Techniques
- NCERT Solutions Class 11 Chemistry Chapter 13 : Hydrocarbons
- NCERT Solutions Class 11 Chemistry Chapter 14 : Environmental Chemistry
