The value of equilibrium constant of the reaction HI(g) leftharpoons (1)/(2) H_2(g)+(1)/(2) I _2 is 8.0 The equilibrium constant of the reaction H_2(g)+ I _2(g) leftharpoons 2 HI(g) will be

Correct answer: B. (1)/(64) gathered HI(g) leftharpoons (1)/(2) H_2(g)+(1)/(2) I_2(g) K_1= [H_2]^ 1 / 2 [I_2]^ 1 / 2 [HI] H_2(g)+I_2(g) leftharpoons 2 HI(g) K_2= [HI^2. [H_2][I_2] gathered From Eqs (i) and (ii) array l & K_1^2=(1)/(K_2) & K_1=8.0 & K_2=(1)/(K_1^2)=(1)/(8^2)=(1)/(64) array

NEET 2008

Exam: NEET
Year: 2008
Subject: Chemistry
Chapter: Coordination Compounds

NEET 2008 · Chemistry · Coordination Compounds

The value of equilibrium constant of the reaction HI(g)...

The value of equilibrium constant of the reaction

\mathrm{HI}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2}

\mathrm{I}

_2

is 8.0 The equilibrium constant of the reaction

\mathrm{H}_2(g)+

\mathrm{I}

_2(g) \rightleftharpoons 2 \mathrm{HI}(g)

will be

Options

A $\frac{1}{16}$
B $\frac{1}{64}$
C16
D $\frac{1}{8}$

Correct answer

B. $\frac{1}{64}$

Step-by-step solution

\begin{gathered} \mathrm{HI}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(g) \\ \mathrm{K}_1=\frac{\left[\mathrm{H}_2\right]^{1 / 2}\left[\mathrm{I}_2\right]^{1 / 2}}{[\mathrm{HI}]} \\ \mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \\ K_2=\frac{\left[\mathrm{HI}^2\right.}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]} \end{gathered}

From Eqs (i) and (ii)

\begin{array}{l} & K_1^2=\frac{1}{K_2} \\ \because & K_1=8.0 \\ \therefore & K_2=\frac{1}{K_1^2}=\frac{1}{8^2}=\frac{1}{64} \end{array}

Practice Coordination Compounds on Quantrex Academy →

More from Coordination Compounds