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If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately
($\log 4 = 0.60$, $\log 5 = 0.69$)
($\log 4 = 0.60$, $\log 5 = 0.69$)
A
45 minutes
B
60 minutes
C
40 minutes
D
50 minutes
Detailed Solution
For a first order reaction: $k = \dfrac{2.303}{t}\log\dfrac{a}{a - x}$
When $t = 60$ min, $x$ = 60 percent of $a$:
$k = \dfrac{2.303}{60}\log\dfrac{100}{100 - 60} = \dfrac{2.303}{60}\log\dfrac{100}{40}$
$\log\dfrac{100}{40} = \log 10 - \log 4 = 1 - 0.60 = 0.40$
$k = \dfrac{2.303 \times 0.40}{60} = 0.0153$ min$^{-1}$
Now for 50 percent completion: $t_{1/2} = \dfrac{2.303}{0.0153}\log\dfrac{100}{100 - 50} = \dfrac{2.303}{0.0153} \times \log 2$
$t_{1/2} = \dfrac{2.303}{0.0153} \times 0.3010$
$t_{1/2} = 45.31$ min, i.e. approximately 45 minutes
When $t = 60$ min, $x$ = 60 percent of $a$:
$k = \dfrac{2.303}{60}\log\dfrac{100}{100 - 60} = \dfrac{2.303}{60}\log\dfrac{100}{40}$
$\log\dfrac{100}{40} = \log 10 - \log 4 = 1 - 0.60 = 0.40$
$k = \dfrac{2.303 \times 0.40}{60} = 0.0153$ min$^{-1}$
Now for 50 percent completion: $t_{1/2} = \dfrac{2.303}{0.0153}\log\dfrac{100}{100 - 50} = \dfrac{2.303}{0.0153} \times \log 2$
$t_{1/2} = \dfrac{2.303}{0.0153} \times 0.3010$
$t_{1/2} = 45.31$ min, i.e. approximately 45 minutes