# Calculate the value of k for this reaction based on the following data collected in three different experiments:

$\begin{array} {cccc} \text{Experiment} & [A]\; \text{in M} & [B]\; \text{in M} & \text{Initial Rate in Ms}^{-1}\\ 1 & 0.185 & 0.133 & 3.35 \times 10^{-4}\\ 2 & 0.185 & 0.266 & 1.35 \times 10^{-3}\\ 3 & 0.370 & 0.133 & 6.75 \times 10^{-4}\\ \end{array}$

$\begin{array}{1 1} 0.102\;M^{-2}\;s^{-1} \\ 0.408 M^{-2}s^{-1} \\ 0.204\;M^{-2}s^{-1} \\ 0.816M^{-2}s^{-1} \end{array}$

Answer: $0.102\;M^{-2}s^{-1}$
Given $\begin{array} {cccc} \text{Experiment} & [A] & [B] & \text{Initial Rate in Ms}^{-1}\\ 1 & 0.185 & 0.133 & 3.35 \times 10^{-4}\\ 2 & 0.185 & 0.266 & 1.35 \times 10^{-3}\\ 3 & 0.370 & 0.133 & 6.75 \times 10^{-4}\\ \end{array}$
The rate of the reaction $= k [A]^ x [B]^ y$
Comparing Experiments 1 and 3, With [B] constant, as [A] is doubled, rate is doubled. Therefore rate is first order with respect to [A] $\rightarrow$ x = 1
Compare experiments 1 and 2: With [A] constant, as [B] is doubled, rate increases by 4 times. Therefore rate is second order with respect to [B] $\rightarrow$ y = 2
$\Rightarrow$ Rate $= k[A][B]^2$
To calculate $k$, Insert data from any experiment into the rate equation:
$\Rightarrow 3.35 \times 10^{-4} = k \times 0.185 \times 0.133^2 \rightarrow k = 0.102\;M^{-2}s^{-1}$