| [A] For 2-butene \[{{\operatorname{CH}}_{3}}CH=CHC{{H}_{3}}\,K({{S}^{-1}})\] |
| \[K({{S}^{-1}})={{10}^{13.8}}\,\exp {{\,}^{-263.5\,\,}}{{^{KJ\,}}^{mol{{\,}^{-1}}/RT}}\] |
| [B] For 2-butene nitrile \[C{{H}_{3}}CH=CH\cdot CN\] |
| \[K({{S}^{-1}})={{10}^{11}}\,\,\exp {{\,}^{-214.5\,\,}}{{^{KJ\,}}^{mol{{\,}^{-1}}/RT}}\] |
| The temperature at which K = K? |
A) 913.87 K
B) 533.43 K
C) 1000.02 K
D) 407.05 K
Correct Answer: A
Solution :
\[\because \,\,\,\,\,\,\,K=K'\] \[{{10}^{13.8}}{{\exp }^{-263.5\,KJ\,mo{{l}^{-1}}/RT}}={{10}^{11}}={{10}^{11}}{{\exp }^{-214.5\,KJ}}\] \[mo{{l}^{-1}}/RT\] or \[{{10}^{13.8-11}}=\exp \frac{[(-214.5)-(-263.5)]\,KJ\,mo{{l}^{-1}}}{8.314\times {{10}^{-3}}\,KJ\,mo{{l}^{-1}}\,{{K}^{o}}\,\times T}\] or \[630.957= exp exp\,\frac{5.893\times {{10}^{3}}}{T}\] \[\therefore \,\,\,\,\,\,2.303\,\,log\,\,630.957\,\,=\,\,\frac{5.893\times {{10}^{3}}}{T}\] \[\therefore \,\,\,\,\,\,\,\,T=\frac{5.893\times {{10}^{3}}}{2.303\times 2.8}=913.87\,\,K\]
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