Question

Which function of [X] plotted against time, will give a straight line for a second order reaction, $X\to \text{Product}$ ?
(A) [X]
(B) ${{[X]}^{2}}$
(C) ln[X]
(D) $\dfrac{1}{[X]}$

Answer 1

Hint: The general equation for a graph to be a straight line is y = mx + c in which m is a slope, x and y are the values of quantities represented in respective axis and c is a constant. We can use a similar equation for a second order reaction to find the function.

Complete step by step answer:
We know that the sum of powers of the concentration of the reactants in the rate law expression will be two. So, we can summarize the rate law expression of a second order reaction as below.
     \[-\dfrac{dx}{dt}=k{{[X]}^{2}}\] …..(1)

Now, in order to identify which function of X will give a straight line against time, we need to find out when a straight line results in a graph.
- If y is represented on y-axis and x is represented on x-axis of the graph, then we can say that if
$y=mx+c$, then the graph will be a straight line. Here, m is a slope and c is a constant.
- Here, we will need to convert equation (1) into the form of the straight line equation. So, let’s convert it.

We are given that \[-\dfrac{dx}{dt}=k{{[X]}^{2}}\]
Now, we will need to do integration of X from the limits ${{x}_{0}}$ to x.
So,we can write that \[-\dfrac{dx}{{{[X]}^{2}}}=kdt\]
     \[\int\limits_{{{x}_{0}}}^{x}{-\dfrac{dx}{{{[X]}^{2}}}}=\int\limits_{0}^{t}{kdt}\]
Thus, $\int\limits_{{{x}_{0}}}^{x}{-\dfrac{1}{X}}=k\int\limits_{0}^{t}{dt}$
So, we can write that $\dfrac{1}{X}-\dfrac{1}{{{X}_{0}}}=kt$ …….(2)
We can also write equation (2) as $\dfrac{1}{[X]}=kt+\dfrac{1}{[{{X}_{0}}]}$
This equation is similar to straight line equation y=mx+c where $\dfrac{1}{[X]}$ is y, k is a slope, t is time and $\dfrac{1}{[{{X}_{0}}]}$ is a constant.

Thus, from a given mathematical operation, we can conclude that $\dfrac{1}{[X]}$ will give a straight line against time in second order reaction.

So, the correct answer is “Option D”.

Note: Remember that as the second order reaction depends upon the power of concentration of reactants which is two, that does not mean that ${{[X]}^{2}}$ will be the function if that reaction gives a straight line.