A) 0.25 atm
B) 0.38 atm
C) 0.46 atm
D) 0.56 atm
Correct Answer: B
Solution :
\[{{H}_{2}}(g)+S(s){{H}_{2}}S(g);\] \[{{K}_{c}}=6.8\times {{10}^{-2}}\] Initial 0.20 0 Atequiation \[(0.20-x)~\] \[x\] \[{{K}_{c}}=6.8\times {{10}^{-2}}=\frac{[{{H}_{2}}S]}{{{H}_{2}}}\] \[6.8\times {{10}^{-2}}=\frac{x}{0.20-x}\] \[x=1.273\times {{10}^{-2}}mol\,{{L}^{-1}}\] \[{{p}_{{{H}_{2}}S}}=\left( \frac{n}{v} \right)RT\] \[=(1.273\times {{10}^{-2}})\times 0.0821\times 363\] \[=0.38\text{ }atm\]You need to login to perform this action.
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