The reaction ${C H}_{3} - {C H}_{2} - {N O}_{2} + {O H}^{-} ⟶ {C H}_{3} - \oplus C H - {N O}_{2} + H_{2} O$ obeys the rate law for pseudo-first-order kinetics in the presence of a large excess of hydroxide ion. If $1$ % of nitroethane undergoes reaction in half a minute when the reactant concentration is $0.002 M$ . What is the pseudo-first-order rate constant?
( $log 99 = 1.996$ )

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Solution

1 per cent of nitroethane undergoes reaction in half a minute when the reactant concentration is $0.002 M$ .

Amount of nitroethane reacted = $\frac{0.002}{0.01} = 0.00002$

The expression for the rate constant is $k = \frac{2.303}{t} log \frac{a}{(a - x)} = \frac{2.303}{0.5} log \frac{0.002}{(0.002 - 0.00002)} = 2 \times 10^{- 2} m i n^{- 1}$ .

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1 per cent of nitroethane undergoes reaction in half a minute when the reactant concentration is 0.002M.Amount of nitroethane reacted = 0.0020.01=0.00002The expression for the rate constant is k=2.303tloga(a−x)=2.3030.5log0.002(0.002−0.00002)=2×10−2min−1.