A) \[1.8\times {{10}^{-3}}\]
B) \[3.6\times {{10}^{-3}}\]
C) \[6.0\times {{10}^{-2}}\]
D) \[1.3\times {{10}^{-5}}\]
Correct Answer: C
Solution :
\[2S{{O}_{2}}(g)+{{O}_{2}}2S{{O}_{3}}(g)\] Equilibrium constant for this reaction, \[K=\frac{{{[S{{O}_{3}}]}^{2}}}{{{[S{{O}_{2}}]}^{2}}[{{O}_{2}}]}\] ...(i) \[S{{O}_{3}}(g)S{{O}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\] Equilibrium constant for this reaction, \[K'=\frac{[S{{O}_{2}}]{{[{{O}_{2}}]}^{\frac{1}{2}}}}{[S{{O}_{3}}]}\] ?(ii) On squaring Eq. (ii) both sides, we have \[{{(K')}^{2}}=\frac{{{[S{{O}_{2}}]}^{2}}[{{O}_{2}}]}{{{[S{{O}_{3}}]}^{2}}}\] \[=\frac{1}{K}\] \[=\frac{1}{278}\] \[\therefore \] \[K'=\sqrt{\frac{1}{278}}\] \[=\sqrt{0.003597}\] \[=5.99\times {{10}^{-2}}\] \[\approx 6\times {{10}^{-2}}\]
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