Question

Calculate the time (in minutes) required for 93.75% decomposition of H2O2 at 450°C if half life for decomposition of H2O2 is 10.17 min at 450°C

Answer 1

40.68

Answer 2

The decomposition of H2O2 is a first-order reaction.

For a first-order reaction, the fraction of reactant remaining after 'n' half-lives is given by: $\frac{[A]_t}{[A]_0} = \left(\frac{1}{2}\right)^n$

Given that 93.75% of H2O2 has decomposed, the percentage of H2O2 remaining is: $100\% - 93.75\% = 6.25\%$

As a fraction, this is: $\frac{[A]_t}{[A]_0} = \frac{6.25}{100} = \frac{1}{16}$

Now, we can find the number of half-lives (n) required: $\left(\frac{1}{2}\right)^n = \frac{1}{16}$ Since $16 = 2^4$, we can write: $\left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^4$ Therefore, $n = 4$ half-lives.

The time required for this decomposition is the number of half-lives multiplied by the half-life period: Time (t) = $n \times t_{1/2}$ Given $t_{1/2} = 10.17$ min. $t = 4 \times 10.17$ min $t = 40.68$ min