Question

Estimate the value of the equilibrium constant at 675 K for each of the following reactions. ΔG∘f for BrCl(g) is −1.0 kJ/mol. The standard molar entropy, S∘, for BrCl(g) is 240.0 J/mol⋅K.

3 significant figures in answer.

A. 2NO2(g)⇌N2O4(g) ΔH∘f for N2O4(g) is 9.16 kJ/mol.

B. Br2(g)+Cl2(g)⇌2BrCl(g) ΔH∘f for BrCl(g) is 14.6 kJ/mol.

Answer 1

Sol :-

(A). Given reaction is :

2NO2 (g) <------------> N2O4 (g)

As,

ΔH⁰_rxn = ΔH⁰_f of products - ΔH⁰_f of reactants

= ΔH⁰_f of N2O4 (g) - 2 x ΔH⁰_f of NO2

= 11.1 KJ - 2 x 33.2 KJ

= 11.1 KJ - 66.4 KJ

= - 55.3 KJ

Also,

ΔS⁰_rxn = S⁰_Products - S⁰_Reactants

= S⁰ of N2O4 - 2 x S⁰ of NO2

= 304.4 J/K - 2 x 240.1 J/K

= -175.8 J/K

From Gibb's-Helmholtz equation :

ΔG⁰_rxn = ΔH⁰_rxn - TΔS⁰_rxn

= -55.3 KJ - (298 K).(-175.8 x 10^-3 KJ/K)

= -55.3 KJ +  52.4 KJ

= - 2.90 KJ

Also,

ΔG⁰_rxn = -2.303RT log K

log K = -ΔG_rxn / 2.303RT

log K = - (-2.90 KJ) / (2.303 x 8.314 x 10^-3 KJ.K^-1mol^-1 x 675 K)

log K = 0.224

K = antilog (0.224)

K = 1.67

Hence, equilibrium constant = K = 1.67

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(B). Given reaction is : Br2 (g) + Cl2 (g) <--------------> 2 BrCl (g)

ΔH⁰_rxn = 2 x ΔH⁰_f of BrCl (g) - [ ΔH⁰_f of Br2 (g) + ΔH⁰_f of Cl2 (g)

= 2 x 14.6 KJ - [ 0 KJ + 0 KJ]

= 29.2 KJ

Similarly ,

ΔS⁰_rxn = 2 x S⁰ of BrCl (g) - [ S⁰ of Br₂ (g) + S⁰ of Cl2 (g) ]

= 2 x 240 J/K - [ 245.5 J/K + 223.1 J/K]

= 480 J/K - 468.6 J/K

= 11.4 J/K

Now, ΔG⁰_rxn = 29.2 KJ - (298 K x 11.4 x 10^-3 J/K)

= 29.2 KJ - 3.397 KJ

= 25.8 KJ

log K = - 25.8 KJ / (2.303 x 8.314 x 10^-3 KJK^-1mol^-1 x 675 K)

K = antilog (- 1.996)

K = 1.00 x 10^-2

Hence, equilibrium constant = 1.00 x 10^-2