Question

At a certain temperature, the following reactions have the equilibrium constants as shown below :
$S\left(s\right)+O_2\left(g\right)\rightleftharpoons {SO}_2\left(g\right);K_{c1}=5\times {10}^{52}$
$2S\left(s\right)+3O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right);K_{c2}={10}^{29}$
What is the equilibrium constant, $K_c$ for the reaction at the same temperature ?
$2{SO}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$

• A. $2.5\times {10}^{76}$
• B. $4\times {10}^{23}$
• C. $4\times {10}^{-77}$
• D. $Noneofthese$

Answer 1

The correct option is A $2.5\times {10}^{76}$

Equilibrium reactions depends upon the reaction equations
$S_{\left(S\right)}+O_{2\left(g\right)}\rightleftharpoons S{O}_{2\left(g\right)};K_C=5\times {10}^{52}$
$2S_{\left(S\right)}+{3O}_{2\left(g\right)}\rightleftharpoons 2S{O}_{2\left(g\right)};K_C={10}^{29}$
The $K_C$ for the reaction is calculated as,
$2{SO}_{2\left(g\right)}+O_{2\left(g\right)}\rightleftharpoons {2SO}_{3\left(g\right)}$
By the equation,
$2S+3O_2\rightleftharpoons {2SO}_3$
$K_C^{\prime \prime }=\frac{{\left[{SO}_3\right]}^2}{{\left[O_2\right]}^3}$
By reversing the first eqaution and doubling the coefficient $K_C^{,}$

${(K}_C^{,}{)}^{-1}$
Multiply both side of the equation by K,

$\Rightarrow 2.5\times {10}^{76}$