Question

The half life for a first order reaction is 5 × 10⁴ s. What percentage of the initial reactant will react in 2 hours?

Answer 1

$ASSALAMUALAIKUM$

In case of a first order reaction the amount of reactant left after n half-lives is equal to a/2^n . Here , a= initial amount of reactant.

So the amount reacted is a(1–1/2^n) and hence the percentage is 100 (1–1/2^n)% .

So, half-life = 5x 10^4–4 s and total time is 2 hours so n= 1.44 x 10^7 .

Now here 1/2^n is 6.94 x 10^-8 so 1–1/2^n is 0.9999999306 and so the percentage is
=99.9 % .

$INSHAALLAH it will help you!$

$\mathbb{\huge{A.K.}}$

Answer 2

Answer:

Here , a= initial amount of reactant. So the amount reacted is a(1–1/2^n) and hence the percentage is 100 (1–1/2^n)% . So, half-life = 5x 10^4–4 s and total time is 2 hours so n= 1.44 x 10^7 . =99.9 %

JAI SHRI RAM