Question

After 59.0 min, 59.0 min, 43.0 % 43.0% of a compound has decomposed. What is the half‑life of this reaction assuming first‑order kinetics?

Answer 1

Answer: The half‑life of this reaction assuming first‑order kinetics is 72.7 minutes

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  =100

a - x = amount left after decay process  

a) for completion of 43 % of reaction  

$t=\frac{2.303}{k}\log\frac{100}{100-43}$

$59.0=\frac{2.303}{k}\times 0.244$

$k=9.53\times 10^{-3}min^{-1}$

b) 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.

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$t_{\frac{1}{2}}=\frac{0.693}{9.53\times 10^{-3}min^{-1}}=72.7min$

The half‑life of this reaction assuming first‑order kinetics is 72.7 minutes