A) 3 : 1
B) 1 : 9
C) 36 : 1
D) 1 : 1
Correct Answer: C
Solution :
From equation, \[\begin{matrix} x & & Y+Z \\ 1 & 0 & 0 \\ (1-\alpha ) & \alpha & \alpha \\ \end{matrix}\begin{matrix} {} \\ \text{Initial}\,\text{mole} \\ \text{mole}\,\text{at}\,\text{equilibrium} \\ \end{matrix}\] \[{{K}_{{{p}_{1}}}}=\frac{{{p}_{y}}\times {{p}_{z}}}{{{p}_{x}}}\] \[=\frac{\left[ \frac{\alpha \times {{p}_{1}}}{1+\alpha } \right]\left[ \frac{\alpha \times {{p}_{1}}}{1+\alpha } \right]}{\left[ \frac{1-\alpha }{1+\alpha } \right]{{p}_{1}}}\] \[{{K}_{{{p}_{1}}}}=\frac{{{\alpha }^{2}}{{p}_{1}}}{1-{{\alpha }^{2}}}\] ?(i) From equation \[\begin{matrix} A & & 2B \\ 1 & {} & 0 \\ (1-\alpha ) & {} & 2\alpha \\ \end{matrix}\begin{matrix} {} \\ \text{Initial}\,\text{mole} \\ \text{mole}\,\text{at}\,\text{equilibrium} \\ \end{matrix}\] \[{{K}_{{{p}_{2}}}}=\frac{{{\left[ \frac{2\alpha }{1+\alpha }.{{p}_{2}} \right]}^{2}}}{\left[ \frac{1-\alpha }{1+\alpha } \right]{{p}_{2}}}=\frac{4{{\alpha }^{2}}{{p}_{2}}}{1-{{\alpha }^{2}}}\] ?(ii) From Eqs (i) and (ii) \[\Rightarrow \]\[\frac{9}{1}=\frac{{{p}_{1}}}{4{{p}_{2}}}\]\[\Rightarrow \]\[\therefore \]\[\frac{{{p}_{1}}}{{{p}_{2}}}=\frac{36}{1}\]
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