Question

The Keq for the equilibrium below is 7.52 ⋅ 10-2 at 480.0 °C. 2Cl2 (g) + 2H2O (g) ⇌ 4HCl (g) + O2 (g) What is the value of Keq at this temperature for the following reaction? 2HCl (g) + 1/2O2 (g) ⇌ Cl2 (g) + H2O (g) (A) 0(B) 3.76×10−2(C) 3.65(D) 7.52 ⋅ 10-2(E) 0.274

Answer 1

Answer:

(C) 3.65

Explanation:

The expression for the Keq for the given reaction is:

$2Cl_2_{(g)}+2H_2O_{(g)}\rightleftharpoons 4HCl_{(g)}+O_2_{(g)}$

$K_{eq}=\frac {[HCl_{(g)}]^4[O_2_{(g)}]}{[Cl_2_{(g)}]^2[H_2O_{(g)}]^2}=7.52\times 10^{-2}$

The expression for the K'eq for the given reaction is:

$2HCl_{(g)}+\frac {1}{2}O_2_{(g)}\rightleftharpoons Cl_2_{(g)}+H_2O_{(g)}$

$K'_{eq}=\frac {[Cl_2_{(g)}][H_2O_{(g)}]}{[HCl_{(g)}]^2[O_2_{(g)}]^{\frac {1}{2}}}$

Reciprocal of keq and taking square root of it, we get K'eq

So,

$K'_{eq}=\frac {[Cl_2_{(g)}][H_2O_{(g)}]}{[HCl_{(g)}]^2[O_2_{(g)}]^{\frac {1}{2}}}=\sqrt {\frac {1}{K_{eq}}}$

$K'_{eq}=\frac {[Cl_2_{(g)}][H_2O_{(g)}]}{[HCl_{(g)}]^2[O_2_{(g)}]^{\frac {1}{2}}}=\sqrt {\frac {1}{7.52\times 10^{-2}}}=3.65$