Consider the following gaseous equilibria with equilibrium constant \[{{k}_{1}}\] and \[{{k}_{2}}\] respectively. |
\[S{{O}_{2}}\left( g \right)+\frac{1}{2}{{O}_{2}}\left( g \right)S{{O}_{3}}\left( g \right)\] |
\[2S{{O}_{3}}\left( g \right)2S{{O}_{2}}\left( g \right)+{{O}_{2}}\left( g \right)\] |
The equilibrium constant is related as: |
A) \[2{{K}_{1}}=K_{2}^{2}\]
B) \[K_{1}^{2}=\frac{1}{{{K}_{2}}}\]
C) \[K_{2}^{2}=\frac{1}{{{K}_{1}}}\]
D) \[{{K}_{2}}=\frac{2}{K_{1}^{2}}\]
Correct Answer: B
Solution :
For the reaction; \[S{{o}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)S{{O}_{3}}(g)\] | |||
Equilibrium constant, \[{{K}_{1}}=\frac{\left[ S{{O}_{3}} \right]}{\left[ S{{O}_{2}} \right]{{\left[ {{O}_{2}} \right]}^{1/2}}}\] | ?.. (i) | ||
For the reaction, \[2S{{O}_{3}}(g)2S{{O}_{2}}(g)+{{O}_{2}}(g)\] | |||
Equilibrium constant, \[{{k}_{2}}=\frac{{{\left[ S{{O}_{2}} \right]}^{2}}\left[ {{O}_{2}} \right]}{{{\left[ S{{O}_{3}} \right]}^{2}}}\] | ??. (ii) | ||
On squaring both sides in Eq.(i), we get | |
\[k_{1}^{2}=\frac{{{\left[ S{{O}_{3}} \right]}^{2}}}{{{\left[ S{{O}_{2}} \right]}^{2}}\left[ {{O}_{2}} \right]}\] | ?.. (iii) |
Eqs. (ii) \[\times \] (iii), we get | |
\[K_{1}^{2}\times {{K}_{2}}=1\] | |
\[{{K}_{2}}=\frac{1}{K_{1}^{2}}\Rightarrow K_{1}^{2}=\frac{1}{{{K}_{2}}}\] |
You need to login to perform this action.
You will be redirected in
3 sec