The rate of a first order reaction is $0.04 m o l e$ ${l i t r e}^{- 1} s^{- 1}$ at $10$ minutes and $0.03$ mol ${l i t r e}^{- 1} s^{- 1}$ at $20$ minutes after initiation. Find the half life of the reaction.

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Solution

We know that $k = \frac{2.303}{t_{2} - t_{1}} log \frac{R_{1}}{R_{2}}$
where $R_{1} = 0.04$ mol $L^{_{1}} s^{- 1}$ , $R_{2} = 0.03$ mol $L^{_{1}} s^{- 1}$
$t_{2} = 20 m i n$ , $t_{1} = 10 m i n$
So, $k = \frac{2.303}{20 - 10} log \frac{0.04}{0.03} = \frac{2.303}{10} log \frac{4}{3}$
$k = 0.0285 s^{- 1}$
Now , $t_{\frac{1}{2}} = \frac{0.693}{k} = \frac{0.693}{0.0285} = 24.315 s$

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The rate of a first order reaction is 0.04mole litre−1s−1 at 10 minutes and 0.03 mol litre−1s−1 at 20 minutes after initiation. Find the half life of the reaction.

We know that k=2.303t2−t1logR1R2where R1=0.04 mol L1s−1 , R2=0.03 molL1s−1t2=20min , t1=10minSo, k=2.30320−10log0.040.03=2.30310log43k=0.0285s−1Now , t12=0.693k=0.6930.0285=24.315s

The rate of a first order reaction is $0.04 m o l e$ ${l i t r e}^{- 1} s^{- 1}$ at $10$ minutes and $0.03$ mol ${l i t r e}^{- 1} s^{- 1}$ at $20$ minutes after initiation. Find the half life of the reaction.