NIOS Spontaneity of Chemical Reactions Intext Questions With Answer
National Institute of Open Schooling (NIOS)
Sr. Secondary, Module-4
Lession-10: NIOS Spontaneity of Chemical Reactions Intext Questions With Answer
INTEXT QUESTIONS 10.1
1. The enthalpy change for the transition of ice to liquid water at 273 K is 6.02 kJ mol–1. Calculate the entropy change for the process.
Answer: We know that Gibbs Helmholtz equation-ΔG = ΔH − TΔS
where, ΔH is Change in enthalpy
ΔS is Change in entropy
T is Temperature in kelvin (K)
As ΔG = 0
so from Gibbs Helmholtz equation, ΔH = TΔS
or, ΔS = ΔH/T
or, ΔS = 6.02 x 103Jmol−1/273K
or, ΔS = 22.051Jmol−1 or, Entropy change (ΔS) for given transition is 22.051Jmol−1
2. Arrange the following systems in the order of increasing randomness,
(i) 1 mol of gas A
(ii) 1 mol of solid A
(iii) 1 mol of liquid A
Answer: Order of increasing randomness-(ii) 1 mol of solid A < (iii) 1 mol of liquid A < (i) 1 mol of gas A
3. Indicate whether you would expect the entropy of the system to increase or decrease
(a) 2SO2(g) + O2(g) → 2SO3(g)
(b) N2(g) + 3H2(g) → 2NH3(g)
(c) O2(g) → 2O(g)
Answer: a. Decreased b. Decreases c. Increased.
INTEXT QUESTIONS 10.2
1. Determine whether the following reaction
CCl4 (l) + H2 (g) → HCl (g) +CHCl3 (l)
is spontaneous at 298 K if ΔrH = 91.35 kJ mol–1 and ΔrS = 41.5 JK–1 mol–1 for this reaction.
Answer: ΔG = –103.7 kJ. Therefore, the reaction is spontaneous.2. Which of the following conditions would predict a process that is always spontaneous?
(i) ΔH > 0, Δ S > 0(ii) Δ H > 0, Δ S < 0
(iii) Δ H < 0, Δ S > 0
(iv) Δ H < 0, Δ S < 0
Answer: iii. Δ H < 0, Δ S > 0
INTEXT QUESTIONS 10.3
1. What is the relationship between the standard Gibbs energy change and the equilibrium constant of the reaction
Answer: ΔGo = – RT lnKOr, ΔGo = – 2.303 RT log K
2. The standard Gibbs energy change for the reaction
CO (g) + 2H2(g) ⎯⎯→ CH3OH (l)
at 298 K is – 24.8 kJ mol−1. What is the value of the equilibrium constant at298 K
Answer:
Temperature(T) = 298KStandard Gibbs energy change(ΔGo) = −24.8KJ
Gas constant(R) = 0.008314kJmol−1K−1
Equilibrium constant(K) = ?
ΔGo = −RTInK
Or, lnK = ΔGo / − RT
Putting all the values of ΔGo, R and T we get-
logK = −24.8 / (−298 × 0.008314 × 2.303)
or, logK = 4.346418
or, K = 22203.20
or, K = 2.2 × 104