A) 0.074
B) 0.0533
C) 0.177
D) 0.088
Correct Answer: D
Solution :
\[\begin{matrix} {} & 2C{{O}_{2}}\rightleftharpoons & 2CO+ & {{O}_{2}} \\ \text{at}\,\text{equilibrium,} & PC{{O}_{2}} & PCO & P{{O}_{2}} \\ {} & =0.6\,atm & =0.4\,atm & =0.2\,atm \\ \end{matrix}\] \[\therefore \] \[{{K}_{p}}=\frac{{{[CO]}^{2-}}[{{O}_{2}}]}{{{[C{{O}_{2}}]}^{2}}}=\frac{{{(0.4)}^{2}}\times (0.2)}{{{(0.6)}^{2}}}\] \[=0.088\]You need to login to perform this action.
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