A) ¾
B) 1
C) 4
D) 1/3
Correct Answer: B
Solution :
| [b] \[2{{A}_{2}}(g)\rightleftharpoons {{A}_{4}}(g)\] |
| \[3P-x-y\text{ }x/2\] |
| \[{{A}_{2}}+2C\rightleftharpoons {{A}_{2}}{{C}_{2}}\] |
| \[3P-x-y\text{ }P-2y\text{ }y-z\] |
| \[{{A}_{2}}{{C}_{2}}\rightleftharpoons 2AC\] |
| \[y-z2z\] | |
| \[\frac{{{P}_{{{A}_{4}}}}}{P_{{{A}_{2}}}^{2}}=K{{P}_{1}}\Rightarrow {{\left( {{P}_{{{A}_{2}}}} \right)}^{2}}=\frac{{{P}_{{{A}_{4}}}}}{{{K}_{{{P}_{1}}}}}=\frac{1/2}{2/81}=\frac{81}{4}\] | |
| \[\Rightarrow \] \[{{P}_{{{A}_{2}}}}=\frac{9}{2}\text{ }atm\text{ }\] | |
| \[\Rightarrow \] \[3P-x-Y=9/2\] | ??.?(i) |
| \[\Rightarrow \] \[y=3P-x-\frac{9}{2}\] | |
| Also given \[{{P}_{{{A}_{2}}}}=\frac{1}{2}\] | |
| \[\Rightarrow \] \[\frac{x}{2}=\frac{1}{2}\] | ???(ii) |
| \[\Rightarrow \] \[x=1\text{ }atm\] | |
| Also given \[{{P}_{AC}}=\frac{1}{2}\Rightarrow 2z=1/2\] | ??..?(iii) |
| \[\Rightarrow \] \[\text{z =1/4 atm}\] |
| \[{{P}_{total}}=3P-x-y+x/2+P-2y+y-z+2z\] |
| \[=4P-x/2-2y+z\] |
| \[\Rightarrow \] \[\frac{27}{4}=4P-\frac{x}{2}-6P+2x+9+z\] |
| On putting the values of x and z, we get P = 2 atm |
| Thus \[y=6-1-\frac{9}{2}=\frac{1}{2}\] |
| Thus \[{{P}_{{{A}_{2}}{{C}_{2}}}}=y-z=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}atm.\] |
| \[\frac{{{n}_{{{A}_{2}}}}}{{{n}_{AC}}}=\frac{{{P}_{{{A}_{2}}}}}{{{P}_{AC}}}=\frac{3P-x-y}{2z}=\frac{9/2}{1/2}=9\] |
| \[{{K}_{P}}=\frac{{{P}_{{{A}_{2}}{{C}_{2}}}}}{P_{A}^{2}C}=\frac{1/4}{{{\left( 1/2 \right)}^{2}}}=1\] |
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