For the second-order decomposition reaction, $ 2 \mathrm{~A} \rightarrow \mathrm{B} $ $ \quad$$ \quad$$ \quad$$ \quad$$ \quad$$ \quad$$\text { rate }=k[\mathrm{~A}]^{2}$
Which of the following relationships yields a straight line plot?

Question

Quizplus · Accepted Answer

The second-order reaction can be rearranged to give the integrated rate law:

$$ \frac{1}{[\mathrm{A}]} = \frac{1}{[\mathrm{A}]_0} + kt $$

Taking the reciprocal of both sides of this equation yields:

$$ [\mathrm{A}] = \frac{[\mathrm{A}]_0}{1 + k[\mathrm{A}]_0t} $$

This equation shows that the concentration of A decreases with time in a hyperbolic fashion. If we plot 1/[A] versus time, we get a straight line with slope = k and intercept = 1/[A]0.

Therefore, the correct answer is (D) 1/[A] versus t.

For the Second-Order Decomposition Reaction 2 A→B 2 \mathrm{~A} \rightarrow \mathrm{B} 2 A→B

For the Second-Order Decomposition Reaction $2 \mathrm{~A} \rightarrow \mathrm{B}$