Question

A first order reaction has k=1.5 x 10^-6 sec^-1 at 240 degree



c. If the reaction is allowed to run for 10 hours what percentage of the initial concentration would have changed into products?

Answer 1

It's your solution.
hope. it helped.

Answer 2

4.8% of the initial concentration would have changed into products.

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  

when reaction is allowed to run for 10 hours

t= 10 hours = $10\times 60\times 60sec=36000sec$

$36000=\frac{2.303}{1.5\times 10^{-6}}\log\frac{100}{(a-x)}$

$\log\frac{100}{(a-x)}=0.023$

$\frac{100}{(a-x)}=1.05$

$(a-x)=95.2$

$(100-x)=95.2$

$x=4.8$

Thus 4.8 % of the initial concentration would have changed into products.

Learn more about first order kinetics

https://brainly.in/question/13866227

https://brainly.com/question/2958705