A) 0.095 atm
B) 0.64 atm
C) 1.1 atm
D) 0.28 atm
Correct Answer: B
Solution :
| \[{{N}_{2}}{{O}_{4}}\,\,\,\,2N{{O}_{2}}\] |
| \[{{K}_{p}}=\frac{P_{N{{O}_{2}}}^{2}}{{{P}_{{{N}_{2}}{{O}_{4}}}}}=\frac{{{(1.1)}^{2}}}{0.28}=4.32\] |
| When vol. is doubled press. Become half & reaction proceeds in forward direction because \[{{Q}_{p}}\propto {{K}_{p}}\] |
| \[{{N}_{2}}{{O}_{4}}2N{{O}_{2}}\] |
| \[\frac{0.28}{2}-P\] \[\frac{1.1}{2}+2P\] |
| \[{{K}_{p}}=\frac{{{(0.55+2P)}^{2}}}{(0.14-P)}=4.32\] \[\Rightarrow p=0.045\] |
| \[{{P}_{N{{O}_{2}}}}\] at new equilibrium \[=0.55+2\times 0.045\]=0.64 atm |
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