A) \[1.2\times {{10}^{-2}}\]
B) \[<1.2\times {{10}^{-2}}\]
C) 83
D) \[>1.2\times {{10}^{-2}}\]
Correct Answer: B
Solution :
\[2{{H}_{2}}S(g)\rightleftharpoons 2{{H}_{2}}(g)+{{S}_{2}}(g)\] \[{{K}_{p}}=1.2\times {{10}^{-12}}\] \[T=1065+273=1338\,K\] \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}\] where \[\Delta n=\]number of moles of products - number of moles of reactants \[\Delta n=3-2=1\] \[R=0.0821\] \[2\times {{10}^{-2}}={{K}_{e}}{{(0.0821\times 1338)}^{1}}\] \[{{K}_{c}}=\frac{1.2\times {{10}^{-2}}}{0.0821\times 1338}=\frac{1.2\times {{10}^{-12}}}{109.8}\] \[=0.0109\times {{10}^{-2}}\] It means that \[{{K}_{c}}<1.2\times {{10}^{-2}}\]You need to login to perform this action.
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