Question

What is the value of the intercept for the graph for log [R] $\to $ time for the first-order reaction?
(a)- $\log \text{ }\!\![\!\!\text{ R}{{\text{ }\!\!]\!\!\text{ }}_{0}}$
(b)- $\ln \text{ }\!\![\!\!\text{ R}{{\text{ }\!\!]\!\!\text{ }}_{0}}$
(c)- $\dfrac{-K}{2.303}$
(d)- $\dfrac{K}{2.303}$

Answer 1

Hint: The intercept of the equation can be calculated by equating the equation of the first-order equation to the straight-line equation $y=mx+c$, in which the c is the intercept.

Complete answer:
The first-order reaction means the rate of the reaction only depends on the concentration of one reactant in the reaction.
We can write the rate of reaction for the first order reaction as:
$k=\dfrac{\ln [R]-\ln {{[R]}_{0}}}{t}$
Where k is the rate constant, t is the time taken, [R] is the concentration of the compound at time t, and ${{[R]}_{0}}$ is the initial concentration of the compound.
This equati0n becomes:
$\ln [R]=\ln {{[R]}_{0}}-kt$
This question has a natural log, so converting it into logarithm we can write:
$\log [R]=\log {{[R]}_{0}}-\dfrac{kt}{2.303}$
Or we can write this equation as:
$\log [R]=-\dfrac{kt}{2.303}+\log {{[R]}_{0}}$
We want to know the intercept for this equation, so we have to equate this equation with the straight-line equation. The straight-line equation is:
$y=mx+c$
This equation is explained as when we draw a graph of an equation between the x-axis and y-axis, m is the slope and c is the intercept.
So the first order reaction can also be written as:
$\log [R]=-\dfrac{k}{2.303}t+\log {{[R]}_{0}}$
So when the graph of log[R] is plotted against time, $-\dfrac{k}{2.303}$ is the slope and $\log {{[R]}_{0}}$ is the intercept.

So the correct answer is an option (a)- $\log \text{ }\!\![\!\!\text{ R}{{\text{ }\!\!]\!\!\text{ }}_{0}}$.

Note:
You may get confused between option (a) and (b), but the correct is option (b) because in the question the graph is between log [R], not ln [R].