# The following reaction was monitored as a function of time: AB -> A + B. A plot of 1/[AB] versus time yields a straight line with slope 5.6xx10^(-2) (M.s)^-1. What is the value of the rate constant (k) for this reaction temperature?

Jul 9, 2016

$\textsf{k = 5.6 \times {10}^{- 2} \textcolor{w h i t e}{x} {\text{mol"^(-1)."l"."s}}^{- 1}}$

#### Explanation:

This looks like a 2nd order reaction:

$\textsf{A B \rightarrow A + B}$

For which the rate of disappearance of $\textsf{A B}$ can be written:

$\textsf{\frac{d \left[A B\right]}{\mathrm{dt}} = - k {\left[A B\right]}^{2}}$

Integrating this gives:

$\textsf{\frac{1}{\left[A B\right]} = \frac{1}{{\left[A B\right]}_{0}} + k t}$

Where $\textsf{{\left[A B\right]}_{0}}$ is the initial concentration and $\textsf{k}$ is the rate constant.

This means a plot of sf((1)/([AB]) against $\textsf{t}$ will give a gradient of $\textsf{k}$ and an intercept of $\textsf{\frac{1}{{\left[A B\right]}_{0}}}$: The graph is a straight line of the form $\textsf{y = m x + c}$ so the rate constant is equal to the gradient of the line:

$\therefore$$\textsf{k = 5.6 \times {10}^{- 2} \textcolor{w h i t e}{x} {\text{mol"^(-1)."l"."s}}^{- 1}}$