At 700 , K, the equilibrium constant K_p for the reaction 2S O_ 3… - Vidyadip

MCQ

At $700\, K$, the equilibrium constant ${K_p}$ for the reaction $2S{O_{3(g)}}$ $\rightleftharpoons$ $ 2S{O_{2(g)}} + {O_{2(g)}}$ is $1.80 \times {10^{ - 3}}$ and $kP_a$ is $14$, ($R = 8.314\, Jk^{-1} \,mol^{-1}$). The numerical value in moles per litre of ${K_c}$ for this reaction at the same temperature will be

✓
$3.09\; \times \;{10^{ - 7}}$ $mol-litre$
B
$5.07\; \times \;{10^{ - 8}}$ $mol-litre$
C
$8.18\; \times \;{10^{ - 9}}$ $mol-litre$
D
$9.24\; \times \;{10^{ - 10}}$ $mol-litre$

✓

Answer

Correct option: A.

$3.09\; \times \;{10^{ - 7}}$ $mol-litre$

(a) $\mathop {2S{O_3}}\limits_2 $ $ \rightleftharpoons $ $\mathop {2S{O_2} + {O_2}}\limits_{\,\,\,\,\,3} $

$\Delta n = 3 - 2 = + 1$; ${K_p} = 1.80 \times {10^{ - 3}}$

${[RT]^{\Delta n}} = {(8.314 \times 700)^1}$

${K_c} = \frac{{{K_p}}}{{{{(RT)}^{\Delta n}}}} = \frac{{1.8 \times {{10}^{ - 3}}}}{{{{(8.314 \times 700)}^1}}}$

$ = 3.09 \times {10^{ - 7}}$ $mole-litre.$

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