Question

For certain chemical reaction $t_{75\%}$ is 1.5 times of its $t_{50\%}$. Order of the reaction is n. For some other reaction $t_{75\%}$ is twice of its $t_{50\%}$. Order of the second reaction is m. Value of (n + m) is _______.

Answer 1

1

Answer 2

For a chemical reaction of order n, the relationship between $t_{75\%}$ and $t_{50\%}$ depends on the order.

For a zero-order reaction (n=0), the integrated rate law is $[A]_t = [A]_0 - kt$.

$t_{50\%}$ is the time when $[A]_t = [A]_0/2$.

$[A]_0/2 = [A]_0 - kt_{50\%} \implies t_{50\%} = \frac{[A]_0}{2k}$.

$t_{75\%}$ is the time when $[A]_t = [A]_0/4$.

$[A]_0/4 = [A]_0 - kt_{75\%} \implies t_{75\%} = \frac{3[A]_0}{4k}$.

The ratio $\frac{t_{75\%}}{t_{50\%}} = \frac{3[A]_0/(4k)}{[A]_0/(2k)} = \frac{3}{2} = 1.5$.

So, for a zero-order reaction, $t_{75\%} = 1.5 \times t_{50\%}$.

For a first-order reaction (n=1), the integrated rate law is $\ln([A]_t) = \ln([A]_0) - kt$.

$t_{50\%}$ is the time when $[A]_t = [A]_0/2$.

$\ln([A]_0/2) = \ln([A]_0) - kt_{50\%} \implies \ln(1/2) = -kt_{50\%} \implies -\ln 2 = -kt_{50\%} \implies t_{50\%} = \frac{\ln 2}{k}$.

$t_{75\%}$ is the time when $[A]_t = [A]_0/4$.

$\ln([A]_0/4) = \ln([A]_0) - kt_{75\%} \implies \ln(1/4) = -kt_{75\%} \implies -2\ln 2 = -kt_{75\%} \implies t_{75\%} = \frac{2\ln 2}{k}$.

The ratio $\frac{t_{75\%}}{t_{50\%}} = \frac{2\ln 2/k}{\ln 2/k} = 2$.

So, for a first-order reaction, $t_{75\%} = 2 \times t_{50\%}$.

For a second-order reaction (n=2), the integrated rate law is $\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$.

$t_{50\%}$ is the time when $[A]_t = [A]_0/2$.

$\frac{1}{[A]_0/2} = \frac{1}{[A]_0} + kt_{50\%} \implies \frac{2}{[A]_0} = \frac{1}{[A]_0} + kt_{50\%} \implies kt_{50\%} = \frac{1}{[A]_0} \implies t_{50\%} = \frac{1}{k[A]_0}$.

$t_{75\%}$ is the time when $[A]_t = [A]_0/4$.

$\frac{1}{[A]_0/4} = \frac{1}{[A]_0} + kt_{75\%} \implies \frac{4}{[A]_0} = \frac{1}{[A]_0} + kt_{75\%} \implies kt_{75\%} = \frac{3}{[A]_0} \implies t_{75\%} = \frac{3}{k[A]_0}$.

The ratio $\frac{t_{75\%}}{t_{50\%}} = \frac{3/(k[A]_0)}{1/(k[A]_0)} = 3$.

So, for a second-order reaction, $t_{75\%} = 3 \times t_{50\%}$.

For the first reaction, $t_{75\%} = 1.5 \times t_{50\%}$. This matches the relationship for a zero-order reaction. Therefore, the order n = 0.

For the second reaction, $t_{75\%} = 2 \times t_{50\%}$. This matches the relationship for a first-order reaction. Therefore, the order m = 1.

The value of (n + m) is 0 + 1 = 1.