Question

A first order reaction is given as $\mathrm{A} \rightarrow$ product. Its integrated equation is:
A. k=$\dfrac{2.303}{{t}} $$log \dfrac{a-x}{{a}}$
B. k=$\dfrac{1}{t} log \dfrac{a}{a-x}$
C. k=$\dfrac{2.303}{t} log \dfrac{a}{a-x}$
D. k=$\dfrac{1}{t} log \dfrac{a-x}{a}$

Answer 1

Hint: We know that the reaction rate can be written in accordance to the two laws. It is known that, the first is the differential rate law and the other is the law of mass action.

Complete step by step answer:
We can assume that the concentration of $\mathrm{A} at time \mathrm{t}$ is denoted by $[\mathrm{A}]_{\mathrm{t}}=a-x$. Therefore, we can say that the initial concentration of $\mathrm{A}$is denoted by $[\mathrm{A}]_{\mathrm{0}}=a$ at time \mathrm{t} and is equals to zero.
We know that the rate law for the consumption of $\mathrm{A}$ for the first order reaction is given by the expression shown as follows.
Rate of consumption of $\mathrm{A}=-\dfrac{\mathrm{d}[\mathrm{A}]}{\mathrm{d} t}=k[\mathrm{A}]$
We know that the instantaneous rate is referred to as the derivative of concentration with respect to time. Thus, when we divide the equation written above both sides by $[\mathrm{A}]$ and multiply by $-\mathrm{dt}$. We get the modified equation as follows.
$\dfrac{d[A]}{[A]}=-k d t$
When we integrate the above equation we get the following result.
$\begin{aligned} \int_{a}^{a-x} \dfrac{\mathrm{d}[\mathrm{A}]}{[\mathrm{A}]} &=-k \int_{0}^{t} \mathrm{d} t \\ \ln (a-x)-\ln (a) &=-k t \\ \ln \left(\dfrac{a-x}{a}\right) &=-k t \\ 2.303 \log \left(\dfrac{a-x}{a}\right) &=-k t \end{aligned}$
Thus, we can conclude that the required expression is the one which is shown below.
$k=\dfrac{2.303}{t} \log \left(\dfrac{a}{a-x}\right)$
Thus, we can say that the correct option is C.

Note:
We can say that in the integrated rate law for the first order reaction, the concentration of the reactant is the exponential function of the time. We can also say that the half-life of the particular chemical reaction is the time at which the amount of reactant gets reduced to half of the initial value.