Question

Given that for a reaction of order , the integrated form of the rate equation is $k=\frac{1}{t(n-1)}\left|\begin{array}{c}\frac{1}{C^{n-1}-\frac{1}{C_0^{n-1}}}\end{array}\right|$
where $C_0$ and C are the values of the reactant concentration at the start and after time t . What is the relationship between $t_{3/4}$ and $t_{1/2}$ where $t_{3/4}$ is the time required for to become $\frac{1}{4}{C}_0$?

• A. $t_{3/4}=t_{1/2}[2^{n-1}+1]$
• B. $t_{3/4}=t_{1/2}[2^{n-1}-1]$
• C. $t_{3/4}=t_{1/2}[2^{n+1}-1]$
• D. $t_{3/4}=t_{1/2}[2^{n+1}+1]$

Answer 1

The correct option is A $t_{3/4}=t_{1/2}[2^{n-1}+1]$

$t_{3/4}=\frac{1}{K(n-1)}.\frac{1}{{\left(\frac{C_0}{4}\right)}^{n-1}}-\frac{1}{C_0^{n-1}}=\frac{1}{K(n-1)}\left(\frac{4^{n-1}-1}{C_0^{n-1}}\right)\frac{t_{3/4}}{t_{1/2}}=\frac{4^{n-1}-1}{2^{n-1}-1}{t}_{3/4}=t_{1/2}(2^{n-1}+1)$