Question

The equilibrium constant, K, for the following reaction is 1.80x102 at 698 K. 2HI(9) =H2(g) +129) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.311 MHI. 4.17x10 - MH, and 4.17x10-2 MIL- What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.174 mol of HI(g) is added to the flask? [HI] = [H2) = [12] =

Answer 1

Explanation:

The equilibrium constant of the reaction = $K_c=1.80\times 10^{-2}$

Initial  concentration of HI = 0.311 M

After addition 0.174 mol of HI(g)

Concentration of HI added = $\frac{0.174 mol}{1 L}=0.174 M$

New concentration of HI = 0.311 M + 0.174 M = 0.485 M

     $2HI\rightleftharpoons H_2+I_2$

Initial  concentration:

0.311 M                $4.71\times 10^{-2} M$           $4.71\times 10^{-2} M$  

At equilibrium:

(0.485 M - x)  $(4.71\times 10^{-2} M+x)$           $(4.71\times 10^{-2} M+x)$  

$K_{eq}=\frac{[H_2][I_2]}{[HI]^2}$

$1.80\times 10^{-2}=\frac{(4.71\times 10^{-2} M+x)\times (4.71\times 10^{-2} M+x)}{ (0.485 M-x)^2}$

$\sqrt{1.80\times 10^{-2}}=\frac{(4.71\times 10^{-2} M+x)}{ (0.485 M-x)}$

$0.1342=\frac{(4.71\times 10^{-2} M+x)}{(0.485 M-x)}$

$0.065087 M-0.1342x=4.71\times 10^{-2} M+x$

$0.065087 M-4.71\times 10^{-2} M=1.1342x$

$x=\frac{0.017987 M}{1.3142}=0.01586 M$

Equilibrium concentrations:

$[HI]=0.485 M-x = 0.485 M - 0.01586 M= 0.46914 M$

$[H_2]=[I_2]=4.71\times 10^{-2} M+x=4.71\times 10^{-2} M+0.01586 M$

$[H_2]=[I_2]=0.06296 M$