93.75% of a $1^{s t}$ order reaction is completed in 120 minutes. If initial concentration of the reactant is 2 mole/lit, the rate of reaction at the end of 120 min is

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$2.31 \times 10^{- 3} M \min^{- 1}$

$2.88 \times 10^{- 3} M \min^{- 1}$

$4.62 \times 10^{- 2} M \min^{- 1}$

$6.93 \times 10^{- 4} M \min^{- 1}$

answer is C.

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Detailed Solution

$t_{93.75 %} = 4 \times t_{50 %} = 120 \min, ∴ t_{50 %} = \frac{120}{4} = 30 \min$

$∴ K = \frac{0.693}{t_{\frac{1}{2}}} = \frac{0.693}{30} = 2.31 \times 10^{- 2} \min^{- 1}$

Amount left after 120 minits is 0.125 mole/liter.

$Rate = K [R] = 2.88 \times 10^{- 3} M \min^{- 1}$

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