## Thermodynamics Quiz-19

Thermodynamics is a very important branch of both chemistry and physics. It deals with the study of energy, the conversion of energy between different forms and the ability of energy to do work. Thermodynamics is an important chapter for JEE Mains, JEE Advance NEET and many others exams. As this chapter is present in both chemistry and physics and there is only a minor difference between them hence it becomes more important topic. It is not a tough topic, it only requires practice. This topic has been given good weightage in all engineering exams. So don't skip this and prepare well. All the best !.

Q1. For a reaction at 25°C enthalpy change(ΔH) and entropy change(ΔS) are −11.7×103 J/mol and −105.0 J/mol/K respectively. The reaction is:
•  Spontaneous
•  Non-spontaneous
•  Instantaneous
•  None of these
Solution:-
ΔG = ΔH − TΔS
∴ ΔG = −11700 − 298 × (−105)
= + 19.59 kJ
Thus, reaction is non-spontaneous.

Q2. The heat of neutralization of strong base and a strong acid is 57 kJ. The heat released when 0.5 mole of HNO3 solution is added to 0.20 mole of NaOH solution, is :
•  11.4 kJ
•  34.7 kJ
•  23.5 kJ
•  58.8 kJ
Solution:-
0.2 mole of HNO3 are neutralized by 0.2 mole of NaOH to give heat = 57 × 0.2 = 11.4 kJ

Q3. Which is the best definition of heat of neutralization?
•  The heat absorbed when one gram molecule of an acid is neutralized by one gram molecule of a base in dilute solution at a stated temperature
•  The heat set free or absorbed when one gram atom of an acid is neutralized by one gram atom of a base at a stated temperature
•  The heat set free or absorbed when a normal solution containing one gram-equivalent of an acid neutralized by a normal solution containing one gram-equivalent of a base at a stated temperature
•  The heat set free when one gram-equivalent of an acid is neutralized by one gram-equivalent of a base in dilute solution at a stated temperature
Solution:-
The heat set free when one gram-equivalent of an acid is neutralized by one gram-equivalent of a base in dilute solution at a stated temperature

Q4. Which one of the following bonds has the highest average bond energy (kcal/mol)?
•  S = O
•  C ≡ C
•  C ≡ N
•  N ≡ N
Solution:-
Bond energy of S = O, C ≡ C, C ≡ N and N ≡ N are 523, 839, 891 and 941 kJ/mol respectively.

Q5. The enthalpy of formation of water from hydrogen and oxygen is −286 kJ/mol. The enthalpy of decomposition of water into hydrogen and oxygen is:
•  −286 kJ/mol
•  −143 kJ/mol
•  +286 kJ/mol
•  +143 kJ/mol
Solution:-
Heat of formation of H2O = − heat of decomposition of water.

Q6. The molar heat capacity of water at constant pressure is 75 J/mol/K. When 1.0 kJ of heat is supplied to 100 g of water which is free to expand the increase in temperature of water is
•  2.4 K
•  3.6 K
•  4.8 K
•  1.2 K
Solution:-
Heat capacity of water per gram = 75/18 = 4.17 J
Q = mst = 100 × 4.17 × t = 417t = 1000
t = 1000/417 = 2.4 K

Q7. Which of the following is correct for an ideal gas :
•  (∂E ∕ ∂T)V = 0
•  (∂E ∕ ∂P)T = 0
•  (∂E ∕ ∂T)P = 0
•  All of these
Solution:-
Internal energy of an ideal gas is function of temperature and thus (∂E ∕ ∂P)T = 0

Q8. The Gibbs energy change for a reversible reaction at equilibrium is:
•  Zero
•  Small positive
•  Small negative
•  Large positive
Solution:-
ΔG = 0 for an equilibrium state.

Q9. Which of the following reaction defines ΔH°f ?
•  Cdiamond + O2(g) → CO2(g)
•  ½H2(g) + ½F2(g) → HF(g)
•  N2(l) + 3H2(g) → 2NH3(g)
•  CO(g) + ½O2(g) → CO2(g)
Solution:-
N2(l) + 3H2(g) → 2NH3(g)

Q10. Standard heat of formation of CH4(g), CO2(g) and water at 25°C are −17.9, −94.1 and −68.3 kcal/mol respectively. Calculate the heat change (in kcal) in the following reaction at 25°C:
CH4(g) +2O2(g) → CO2(g) + 2H2O(l)
•  −144.5
•  −180.3
•  −248.6
•  −212.8
Solution:-
C + 2H2 → CH4;   ΔH = −17.9 kcal......(i)
C + O2 → CO2;   ΔH = −94.1 kcal......(ii)
H2 + ½O2 → H2O;   ΔH = −68.3 kcal......(iii)
eq. [(ii) + 2×(iii)]−(i)
CH4(g) +2O2(g) → CO2(g) + 2H2O(l) ## Want to know more

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