Answer
$5.56\times10^{23}$
Work Step by Step
The key is to figure out how to obtain the object reaction equation for which we need to find $K_c$ by processing the two given reaction equations with given $K_c$ values, and process the $K_c$ values with algebraic method correspondingly.
Step 1: Notice in the object reaction equation $SO_{2}$ has coefficient of 2, and in the given equation (1) with $K_{C}^{′}$ it has coefficient of 1. So, we need multiply the coefficient of $SO_{2}$ in equation (1) by 2; in this way, we multiply the coefficients for all components by 2. As in the $K_c$ expression, the concentrations for all gaseous components are raised to power of their reaction coefficients, we square $K_{C}^{′}$ as well.
S(s) + $O_{2}$(g) $SO_{2}$(g) (1)===> 2S(s) + 2$O_{2}$(g) 2$SO_{2}$(g) (3)
$K_{C}^{′′′}$=$(K_{C}^{′})^2$=$(4.2\times10^{52})^2$=$1.764\times10^{105}$
Step 2: Notice in the object reaction equation $SO_{2}$ is in the reactant side, and in equation (3) we got from step 1 it’s still in the product side. So, we need reverse the product and reactant in equation (3). We switch the product and reactant in $K_{C}^{′′′}$ expression as well, and therefore we take reciprocal of $K_{C}^{′′′}$ we got from step 1.
2S(s) + 2$O_{2}$(g) 2$SO_{2}$(g) ===> 2$SO_{2}$(g) 2S(s) + 2$O_{2}$(g) (4)
$K_{C}^{(4)}$=$\frac{1}{K_{C}^{′′′}}$=$\frac{1}{1.764\times10^{105}}$$\approx$$5.67\times10^{-106}$
Step 3: Now let’s try to directly add up the given equation (2) with $K_{C}^{′′}$ and equation (4) we just got from step 2. By doing that, we can successfully get the object equation:
2$SO_{2}$(g) 2S(s) + 2$O_{2}$(g) (4)
2S(s) +3$O_{2}$(g) 2$SO_{3}$(g) (2)
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2$SO_{2}$(g) + $O_{2}$(g) 2$SO_{3}$(g) -- The object equation
When we add up these two equations, the final $K_{C}$ value of the object equation is the product of $K_{C}$ values of these two equations:
($K_{C}^{(4)}$)$\times$($K_{C}^{′′}$)=($\frac{[O_{2}]^2}{[SO_{2}]^2}$)$\times$($\frac{[SO_{3}]^2}{[O_{2}]^3}$)
=$\frac{[SO_{3}]^2}{[SO_{2}]^{2}[O_{2}]}$
=$K_{C}$ value of the object equation
($K_{C}^{(4)}$)$\times$($K_{C}^{′′}$)=（$5.67\times10^{-106}$）$\times$($9.8\times10^{128}$)$\approx$$5.56\times10^{23}$
