| \[A\to B;\,\,\,\,\,\,\,\,\,\,\,\,\,{{k}_{A}}={{10}^{15}}{{e}^{-2000/T}}\] |
| \[C\to D;\,\,\,\,\,\,\,\,\,\,\,\,\,{{k}_{C}}={{10}^{14}}{{e}^{-1000/T}}\] |
| Temperature T kelvin at which \[(\,{{k}_{A}}={{k}_{C}})\] is: |
A) 1000 K
B) 2000 K
C) (2000/2.303) K
D) (1000/2.303) K
Correct Answer: D
Solution :
| \[{{K}_{A}}={{10}_{15}}{{e}^{-\frac{2000}{T}}};\,{{K}_{B}}={{10}^{14}}{{e}^{-\frac{1000}{T}}}\] |
| \[{{10}^{15}}{{e}^{-}}\frac{2000}{T}={{10}^{14}}{{e}^{-}}\frac{1000}{T}10{{e}^{-1\,}}\frac{2000}{T}={{e}^{-}}\frac{1000}{T}\] |
| \[10={{e}^{+}}\frac{2000-1000}{T}10=e\frac{1000}{T}\ln \,10=\frac{1000}{T}\] |
| \[2.303\log 10=\frac{1000}{T}\,\,\,\,\,\Rightarrow \,\,T=\left[ \frac{1000}{2.303} \right]K\] |
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