| Direction (Q. Nos. 56): 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) 1.78 atm
B) 2.01 atm
C) 3.18 atm
D) 4.33 atm
Correct Answer: A
Solution :
At 300 K \[{{N}_{2}}{{O}_{4}}\,2N{{O}_{2}}\] Initial a 0 At (equi.) \[a(1-0.2)\] \[2\times 0.2a\] = 0.8a = 0.4a At 400 K \[{{N}_{2}}{{O}_{4}}\,2N{{O}_{2}}\] Intial a 0 At (equi.) \[a(1-0.6)\] \[2\times 0.2a\] = 0.4a = 1.2a \[pV=nRT\] \[1\times V=1.2a\times R\times 300\] ?(i) \[p\times V=1.6a\times R\times 400\] ?(ii) Solving Eqs. (i) and (ii) \[p=1.78\,\,atm\,\] at 400 K
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