Question 17.2: Finding K for Reactions Multiplied by a Common Factor or Rev......

Plan We must manipulate reactions (1) and (2) so that they add together to give the target reaction. Reaction (1) must be reversed; the new equilibrium constant (K^′_{c1}) is the reciprocal of K_{c1}. Reaction (2) must be multiplied by 3; the new equilibrium constant (K^′_{c2}) is K_{c2} raised to the third power. We then add the two manipulated reactions and multiply their K_c values to obtain the overall K_c.

Solution Reaction (1) must be reversed since 2NH_3 is a reactant in the target reaction; the K_c value of the reversed reaction is the reciprocal of the original K_c:

Reaction (2) must be multiplied by 3 (in the target reaction I_2 has a coefficient of 3 and HI has a coefficient of 6) so the new K_c value equals the original K_c value raised to the third power:

Adding the two reactions (cancelling common terms):

              (1)     2NH_3(g)  \xrightleftharpoons[]{}  N_2(g)  +  \cancel{3H_2(g)}                                    K^′_{c1}  =  2.92
               (2)     \cancel{3H_2(g)}  +  3I_2(g)  \xrightleftharpoons[]{}  6HI(g)                     K^′_{c2}  =  1.6×10^5
             \overline{2NH_3(g)  +  3I_2(g) \xrightleftharpoons[]{}  N_2(g)  +  6HI(g)}

Multiplying the individual reaction quotients and calculating the overall K_c:

              Q^′_{c1}  ×  Q^′_{c2}  =  \frac{[N_2] \cancel{[H_2]^3}}{[NH_3]^2}  ×  \frac{[HI]^6}{\cancel{[H_2]^3}[I_2]^3}  =  \frac{[N_2] [HI]^6}{[NH_3]^2[I_2]^3}  =  Q_{\text{c(overall)}}
              K_{\text{c(overall)}}  =  K^′_{c1}  ×  K^′_{c2}  =  (2.92)(1.6×10^5)  =  4.7×10^5

Check Round off and check the calculation of the overall K_c value:
              K_c  ≈  (3)(2×10^5)  =  6×10^5