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

$\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:

$\Rightarrow 3.35 \times 10^{-4} = k \times 0.185 \times 0.133^2 \rightarrow k = 0.102\;M^{-2}s^{-1}$

Ask Question

Tag:MathPhyChemBioOther

Take Test

...