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Question

Chemistry Question on Rate of a Chemical Reaction

t1/2t_{1 / 2} for a first order reaction is 10min10\, \min. Starting with 10M10\, M, the rate after 20min20\, \min is

A

0.0693M min1 0.0693\, M\ \min^{-1}

B

0.0693×5Mmin1 0.0693\times 5\,M\, \min^{-1}

C

0.0693×2.5Mmin1 0.0693\times 2.5\,M\,\min^{-1}

D

0.0693×10Mmin1 0.0693\times 10\,M\, \min^{-1}

Answer

0.0693×2.5Mmin1 0.0693\times 2.5\,M\,\min^{-1}

Explanation

Solution

Given, t1/2=10mint_{1 / 2}=10\, \min t=20mint=20\, \min N0=10MN_{0}=10\, M t=nt1/2\Rightarrow t=n t_{1 / 2} 20=n×1020=n \times 10 n=2\therefore n=2 NN0=(12)n\Rightarrow \frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{n} N10=(12)2\Rightarrow \frac{N}{10}=\left(\frac{1}{2}\right)^{2} or N10=14\frac{N}{10} =\frac{1}{4} N=104=2.5M\therefore N =\frac{10}{4}=2.5\, M \therefore Rate =k[A]=k[A] =0.0693×2.5Mmin1=0.0693 \times 2.5\, M\, \min ^{-1}