| Direction (Q. Nos. 58): A given sample of \[{{N}_{2}}{{O}_{4}}\] in a closed shows 20% dissociation in \[N{{O}_{2}}\] at\[{{27}^{o}}C\]and 1 atm. The sample is now heated up to \[{{127}^{o}}C\] and the analysis of the mixture shows 60% dissociation at\[{{127}^{o}}C\]. |
A) 0.165
B) 0.29
C) 0.523
D) 0.625
Correct Answer: A
Solution :
At 300 K \[{{p}_{N{{O}_{2}}}}={{X}_{N{{O}_{2}}}}\cdot {{p}_{T}}\] \[=\frac{0.4\,a}{1.2a}\times 1=0.33\] \[{{p}_{{{N}_{2}}{{O}_{4}}\,}}=\frac{0.8a}{1.2a}\times 1=0.66\] \[\therefore \] \[{{K}_{p}}=\frac{{{({{p}_{N{{O}_{2}}}})}^{2}}}{({{p}_{{{N}_{2}}{{O}_{4}}}})}=\frac{{{(0.33)}^{2}}}{0.66}=0.165\]
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