Question

At a certain temperature, the following reactions have the equilibrium constants as shown below:
$S(s) + {O_2}(g) \rightleftarrows S{O_2}(g);$ ${K_{c1}} = 5 \times {10^{52}}$
$2S(s) + 3{O_2}(g) \rightleftarrows 2S{O_3}(g);$ ${K_{c2}} = {10^{29}}$
What is the equilibrium constant, ${K_c}$ for the reaction at the same temperature? $2S{O_2}(g) + {O_2}(g) \rightleftarrows 2S{O_3}(g)$
(a) 2.5 \times {10^{76}}$$$$$ (b) 4 \times {10^{23}}$(c)$4 \times {10^{ - 77}}$
(d) None of these

Answer 1

The equilibrium constant, K, expresses the relationship between products and reactants of a reaction at equilibrium with respect to a specific unit.

Complete step-by-step solution: (A) It is important to remember that an equilibrium constant is always tied to a specific chemical equation, and if we write the equation in reverse or multiply its coefficients by a common factor, the value of K will change. Fortunately, the rules are very simple:
Writing the equation in reverse will invert the equilibrium expression and
Multiplying the coefficients by a common factor will raise K to the corresponding power.
(B) Notice to get the reaction $2S{O_2}(g) + {O_2}(g) \rightleftarrows 2S{O_3}(g)$
We have to cancel out S from the two given reactions
$S(s) + {O_2}(g) \rightleftarrows S{O_2}(g);$……….. (1)
$2S(s) + 3{O_2}(g) \rightleftarrows 2S{O_3}(g);$…………. (2)
For the cancellation of S, reverse the reaction (1) then the equilibrium constant also get reversed:
$S{O_2}(g) \rightleftarrows S(s) + {O_2}(g)$ $K_{c1}^1 = \dfrac{1}{{5 \times {{10}^{52}}}}$
Notice that we have to multiply the reverse of the reaction (1) with 2, then
The equilibrium constant also get raised to the power 2
$2S{O_2}(g) \rightleftarrows 2S(s) + 2{O_2}(g)$……………. (3) $K_{c1}^{''} = \dfrac{1}{{{{(5 \times {{10}^{52}})}^2}}}$
Notice by adding equation (2) and (3) the final reaction $2S{O_2} + {O_2} \rightleftarrows 2S{O_3}$ is obtained.
(C) The equilibrium constant for the sum of two or more reactions is the product of the equilibrium constants for each of the steps. So the equilibrium constant of reaction (2) and (3) gets multiplied to give the resultant ${K_c}$: ${K_c} = K_{c1}^{''} \times {K_{c2}}$
${K_c} = \dfrac{{{{10}^{29}}}}{{{5^2} \times {{10}^{104}}}} = \dfrac{1}{{25}} \times {10^{ - 75}} = 4 \times {10^{ - 77}}$
${K_c} = 4 \times {10^{ - 77}}$

Hence the correct option is (c).

Note: To determine equilibrium constant for a reaction that is the sum of two or more reactions, add the reactions but multiply the equilibrium constants.