Equilibrium Question 256
Question: At 700 K, the equilibrium constant $ K_{p} $ for the reaction $ 2S{O_{3(g)}} $ ⇌ $ 2S{O_{2(g)}}+{O_{2(g)}} $ is $ 1.80\times {10^{-3}} $ and kPa 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 [AFMC 2001]
Options:
A) $ 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
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Answer:
Correct Answer: A
Solution:
$ \underset{2}{\mathop{2SO_3}}, $ ⇌ $ \underset{,3}{\mathop{2SO_2+O_2}}, $
$ \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.
