| If 60% of a first order reaction was completed in 60 min 50% of the same reaction would be completed in approximately: [AIPMT (S) 2007] |
| (log 4 = 0.60 log 5 = 0.69) |
A) 50 min
B) 45 min
C) 60 min
D) 40 min
Correct Answer: B
Solution :
| [b] \[k=\frac{2.303}{t}{{\log }_{10}}\frac{a}{a-x}\] |
| \[{{k}_{1}}=\frac{2.303}{{{t}_{1}}}\log \frac{{{a}_{1}}}{{{a}_{1}}-{{x}_{1}}}\] |
| \[{{k}_{2}}=\frac{2.303}{{{t}_{2}}}\log \frac{{{a}_{2}}}{{{a}_{2}}-{{x}_{2}}}\] |
| \[{{x}_{1}}=\frac{60}{100}{{a}_{1}},\,{{t}_{1}}=60\] |
| \[{{x}_{2}}=\frac{50}{100}{{a}_{2}},{{t}_{2}}=?\] |
| \[\frac{2.303}{{{t}_{1}}}\log \frac{{{a}_{1}}}{{{a}_{1}}-{{x}_{1}}}=\frac{2.303}{{{t}_{2}}}\log \frac{{{a}_{2}}}{{{a}_{2}}-{{x}_{2}}}\] |
| \[\frac{2.303}{60}\log \frac{a}{\left( {{a}_{1}}-\frac{60}{100}{{a}_{i}} \right)}=\frac{2.303}{{{t}_{2}}}\log \frac{{{a}_{2}}}{\left( {{a}_{2}}-\frac{50}{100}{{a}_{2}} \right)}\] |
| \[\frac{2.303}{60}\log \frac{100\,{{a}_{1}}}{40\,{{a}_{1}}}=\frac{2.303}{{{t}_{2}}}\log \frac{100\,{{a}_{2}}}{50\,{{a}_{2}}}\] |
| \[\frac{1}{60}\log \frac{100}{40}=\frac{1}{{{t}_{2}}}\log \frac{100}{50}\] |
| \[{{t}_{2}}=\frac{60\log 100/50}{\log 100/40}\] |
| \[=\frac{60\,\log \,10-\log \,5)}{(\log \,10-\log 4)}\] |
| \[=\frac{60\,(1-0.69)}{(1-0.60)}=\frac{60\times 0.31}{0.40}\] |
| \[=1.5\times 31=46.5\approx 45\,\min \] |
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