A) 9.84 K
B) 4.92 K
C) 2.45 K
D) 19.67 K
Correct Answer: B
Solution :
[b] \[\log \frac{{{K}_{2}}}{{{K}_{1}}}=\frac{E}{2.303R}\left[ \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right]\] For reaction A - Given,\[\frac{{{K}_{2}}}{{{K}_{1}}}=2,\]\[{{T}_{1}}=300\,\,K,\]\[{{T}_{2}}=310\,\,K\] \[\log 2=\frac{E}{2.303\,R}\left[ \frac{1}{300}-\frac{1}{310} \right]\] ?(i) For reaction B - Given, \[\frac{{{K}_{2}}}{{{K}_{1}}}=2,\]\[E=2E,\]\[{{T}_{1}}=300\,\,K,\]\[{{T}_{2}}=?\] \[\log 2=\frac{2E}{2.303\,R}\left[ \frac{1}{300}-\frac{1}{{{T}_{2}}} \right]\] ?(ii) From equation (i) and (ii), \[\frac{2E}{2.303\,R}\left[ \frac{1}{300}-\frac{1}{{{T}_{2}}} \right]=\frac{E}{2.303\,R}\left[ \frac{1}{300}-\frac{1}{310} \right]\] \[\Rightarrow \]\[2\left[ \frac{1}{300}-\frac{1}{{{T}_{2}}} \right]=\frac{310-300}{300\times 310}\] \[\Rightarrow \]\[{{T}_{2}}=304.92\,\,K\] \[{{T}_{1}}=300\,K,\]\[{{T}_{2}}=304.92\,K\] \[\Delta T={{T}_{2}}-{{T}_{1}}=4.92\,K.\]
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