A) 24.1 min
B) 34.8 min
C) 60.1 min
D) 48.2 min
Correct Answer: A
Solution :
| For first order reaction \[k=\frac{2.303}{t}\log \left( \frac{a}{a-x} \right)\] |
| \[\frac{kt}{2.303}=\log a-\log \,(a-x)\] |
| \[\frac{k{{t}_{1}}}{2.303}=\log a-\log (a-{{x}_{1}})\]at time \[{{t}_{1}}\] |
| \[\frac{k{{t}_{2}}}{2.303}=\log a-\log (a-{{x}_{2}})\]\[{{t}_{2}}\] |
| \[\therefore \]\[\frac{k}{2.303}({{t}_{2}}-{{t}_{1}})=log\left( \frac{a-{{x}_{1}}}{a-{{x}_{2}}} \right)\] |
| Rate, \[{{r}_{1}}=k(a-{{x}_{1}})\] |
| \[{{r}_{2}}=k(a-{{x}_{2}})\] |
| \[\therefore \]\[\frac{(a-{{x}_{1}})}{(a-{{x}_{2}})}=\frac{{{r}_{1}}}{{{r}_{2}}}\]\[\Rightarrow \]\[\frac{0.04}{0.03}=\frac{4}{3}\] |
| \[\frac{k(20-10)}{2.303}=\log \frac{4}{3}\] |
| \[\therefore \]\[k=\frac{2.303}{10}\log \frac{4}{3}=\frac{2.303}{10}\log 1.33\] |
| \[=\frac{2.303\times 0.125}{10}=0.02878\,{{\min }^{-\,1}}\] |
| Also, for 1st order \[{{t}_{1/2}}=\frac{0.693}{k}\] |
| \[{{T}_{50}}=\frac{0.693}{0.02878}=24.1\min \] |
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