Question

if 80% of first order reaction completed in 20 minutes find the out of the time required to complete it 95%​

Answer 1

$t = \frac{2.303}{ \gamma } log_{10}( \frac{a}{a - x} ) \\ \\ 20= \frac{2.303}{ \gamma } log_{10}( \frac{a}{a - \frac{80a}{100} } ) \\ \\ 20 = \frac{2.303}{ \gamma } log_{10}(5) \\ \\ \gamma = \frac{2.303}{20} log_{10}(5) \\ \\ now \\ \\ {t}^{i} = \frac{2.303}{ \gamma } log_{10}( \frac{a}{a - \frac{95a}{100} } ) \\ \\ {t}^{i} = \frac{2.303}{ \frac{2.303}{20} log_{10}(5) } log_{10}(20) \\ \\ {t}^{i} = 20 \times \frac{1.301}{0.699} \\ \\ {t}^{i} = 20 \times0 .861 \\ \\ {t}^{i} = 37.22 \: minutes$

hope it helps.

Answer 2

Answer:

If 80% of the reaction is completed in 20minutes then the portion left will be 100-80=20 hence 0.2part of the original concentration will be left.

If initial concentration is Ao then final concentration A will be 0.2Ao.

Hence, applying the formula to calculate K.

K=1/t * ln(Ao/A).

K= 1/20 *ln(10/2).

K=1/20*ln(5).

Now, to find the time when the reaction is 95% completed hence final concentration is A=0.05Ao.

K=1/t * ln(Ao/A).

K= 1/t *ln(100/5).

K=1/t*ln(20).

So, equating both the equations we get the value of t as.

t = 20*1.86 = 37.2minutes.