Question

Given that for a reaction of order $n$, the integrated form of the rate equation is $k=\frac{1}{t(n-1)}[\frac{1}{C^{n-1}}-\frac{1}{C_0^{n-1}}]$ 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 $C$ to become $1/4C_0$?

• 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 $t_{3/4}=t_{1/2}[2^{n-1}+1]$

for $t_{\frac{3}{4}}$ the concentration $C=\frac{C_0}{4}$ and for $t_{\frac{1}{2}}$ concentration $C^{\prime }=\frac{C_0}{4}$. We can then use these values to form:

$t_{\frac{1}{2}}=\frac{1}{k(n-1)}[\frac{2^{n-1}}{C_0^{n-1}}-\frac{1}{C_0^{n-1}}]=\frac{1}{k(n-1)}\left[\frac{(2^{n-1}-1)}{C_0^{n-1}}\right]\to \left(1\right)$

$t_{\frac{3}{4}}=\frac{1}{k(n-1)}\left[\frac{(2^{2n-2}-1)}{C_0^{n-1}}\right]\to \left(2\right)$

taking $\frac{t_{\frac{3}{4}}}{t_{\frac{1}{2}}}$:

$\frac{t_{\frac{3}{4}}}{t_{\frac{1}{2}}}=2^{n-1}+1\Rightarrow t_{\frac{3}{4}}=t_{\frac{1}{2}}(2^{n-1}+1)$