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

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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

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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 Δ_rS = 41.5 JK–1 mol^–1 for this reaction.

Answer: ΔG = –103.7 kJ. Therefore, the reaction is spontaneous.

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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: ΔG^o = – RT lnK
Or, ΔG^o = – 2.303 RT log K

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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) = 298K
Standard Gibbs energy change(ΔG^o) = −24.8KJ
Gas constant(R) = 0.008314kJmol⁻¹K⁻¹
Equilibrium constant(K) = ?
ΔG^o = −RTInK
Or, lnK = ΔG^o / − RT
Putting all the values of ΔG^o, 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

NIOS Sr. Secondary Chemistry Intext Questions of All Chapters with Answer