Question

For a first order reaction prove that t75% =2t50%?

Answer 1

Explanation:

Hope I was able to solve your doubt

Answer 2

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{50}=\frac{2.303}{k}\log\frac{100}{50}$

$t_{50}=\frac{2.303}{k}\log(2)$

b) for completion of 75 % of reaction

$t=\frac{2.303}{k}\log\frac{100}{100-75}$

$t=\frac{2.303}{k}\log\frac{100}{25}$

$t=\frac{2.303}{k}\times log(4)$

$t_{75}=\frac{2.303}{k}\times log(2)^2$

$t_{75}=\frac{2.303}{k}\times 2\times log(2)$

$\frac{t_{75}}{t_{50}}=\frac{\frac{2.303}{k}\times 2\times log(2)}{\frac{2.303}{k}\times log(2)}$

$frac{t_{75}}{t_{50}}=2$

Thus it is proved that time take for completion of 75% for reaction is 2 times of that of half life.

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