Question

the time for half change in a first order decomposition of a substance A is 60 sec. calculate the rate constant .How much of A will be left after 180 sec​

Answer 1

Answer: 1. $k=0.012sec^{-1}$

2. $frac{1}{8}$ of A will be left after 180 sec​

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 ) to calculate the rate constant

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

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

$k=\frac{0.693}{60}=0.012sec^{-1}$

b) to calculate much of A will be left after 180 sec​

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

$180=\frac{2.303}{0.012}\log\frac{100}{a-x}$

$(a-x)=12.5$

Thus $\frac{1}{8}$ of A will be left after 180 sec​

Answer 2

Here is a pdf to explain and solve the problem in an easier way.

593bc318811815ff94ad5f388dca0753.pdf