Question

In a first order reaction the a/(a- x) was fond to be 8 after 10 minute. The rate constant is

• A. (2.303 x 3 log 2)/10
• B. (2.303 x 3 log 3)/10
• C. 10 x 2.303 x 2 log 3
• D. 10 x 2.303 x 3 log 2

Answer 1

Answer: (A)

(2.303 x 3 log 2)/10

Answer 2

The integrated rate law for a first-order reaction is given by:

$k = \frac{2.303}{t} \log \left(\frac{[A]_0}{[A]_t}\right)$

where $k$ is the rate constant, $t$ is the time, $[A]_0$ is the initial concentration of the reactant, and $[A]_t$ is the concentration of the reactant at time $t$.

In the given problem, we are given that the reaction is first order. The time given is $t = 10$ minutes. The ratio $\frac{[A]_0}{[A]_t}$ is given as $\frac{a}{a-x} = 8$.

Substitute these values into the integrated rate law equation:

$k = \frac{2.303}{10} \log (8)$

We can express $\log(8)$ in terms of $\log(2)$:

$\log(8) = \log(2^3) = 3 \log(2)$

Substitute this back into the equation for $k$:

$k = \frac{2.303}{10} \times 3 \log(2)$

$k = \frac{2.303 \times 3 \log 2}{10}$