Question

At a certain temperature, the following reactions have the equilibrium constant as shown below:

$S\left(s\right)+O_2\left(g\right)\rightleftharpoons S{O}_2\left(g\right);$

$K_C=5\times {10}^{52}$

$2S\left(s\right)+3O_2\rightleftharpoons 2S{O}_3\left(g\right)$

$K_C={10}^{29}$

What is the equilibrium constant $K_C$ for the reaction at the same temperature?

$2S{O}_2\left(s\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$

• A. $4\times {10}^{-77}$
• B. none of these
• C. $2.5\times {10}^{76}$
• D. $4\times {10}^{23}$

Answer 1

The correct option is A $4\times {10}^{-77}$

Using the given equations to derive the desired equation:

$S\left(s\right)+O_2\left(g\right)\rightleftharpoons S{O}_2\left(g\right);K_{c\left(I\right)}=5\times {10}^{52}\dots \left(i\right)$

$2S\left(s\right)+3O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right);K_{c\left(II\right)}={10}^{29}\dots \left(ii\right)$

On reversing equation $\left(i\right)$ and multiplying by $2$,

$2S{O}_2\leftrightharpoons 2S\left(s\right)+2O_2\left(g\right);\dots \left(iii\right)$

$\therefore K_{c\left(III\right)}=\frac{1}{(5\times {10}^{52}{)}^2}$

$\text{So,}{K}_{c\left(III\right)}=(5\times {10}^{52}{)}^{-2}$

Finding the equilibrium constant of required equation.

On adding equations $\left(ii\right)\text{and}\left(iii\right)$,

$2S{O}_2\left(s\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$

$K_c=K_{c\left(II\right)}\times K_{c\left(III\right)}$

$\text{Therefore,}{K}_c=(5\times {10}^{52}{)}^{-2}\times {10}^{29}$

$\Rightarrow K_c=0.04\times {10}^{-104}\times {10}^{29}$

$\Rightarrow K_c=0.04\times {10}^{-74}$

$\Rightarrow K_c=4\times {10}^{-77}$

So, option (C) is the correct answer.

Additional Information

• When reactions are added to get another reaction then the equilibrium constant of the net reaction is the product of equilibrium constant of all the added reactions.
• When reactions are reversed to get another reaction then the equilibrium constant of the new reaction is the inverse of the equilibrium constant of the initial reaction.
• When a reaction is multiplied by a number ‘n’ then the equilibrium constant of the new reaction is the $n^{th}$ power of the initial equilibrium constant.