$\begin{array}{1 1} 5.6 \times 10^{-3}M^{-1} s^{-1} \\ 5.6 \times 10^{-2}M^{-2} s^{-1} \\ 5.6 \times 10^{-3}M^{-2} s^{-1} \\ 5.6 \times 10^{-2}M^{-1} s^{-1} \end{array}$

$\begin{array} {cccc}
\text{Experiment} & [A]\; \text{in M} & [B]\; \text{in M} & \text{Initial Rate in Ms}^{-1}\\
1 & 0.5 & 1.5 & 4.20 \times 10^{-3}\\
2 & 1.5 & 1.5 & 1.26 \times 10^{-2}\\
3 & 3.0 & 3.0 & 5.04\times 10^{-2}\\
\end{array}$

$\begin{array}{1 1} 5.6 \times 10^{-3}M^{-1} s^{-1} \\ 5.6 \times 10^{-2}M^{-2} s^{-1} \\ 5.6 \times 10^{-3}M^{-2} s^{-1} \\ 5.6 \times 10^{-2}M^{-1} s^{-1} \end{array}$

Answer: $ 5.6 \times 10^{-3} M^{-1}s^{-1}$

Given $\begin{array} {cccc} \text{Experiment} & [A]\; \text{in M} & [B]\; \text{in M} & \text{Initial Rate in Ms}^{-1}\\ 1 & 0.5 & 1.5 & 4.20 \times 10^{-3}\\ 2 & 1.5 & 1.5 & 1.26 \times 10^{-2}\\ 3 & 3.0 & 3.0 & 5.04\times 10^{-2}\\ \end{array}$

The rate of the reaction $ = k [A]^ x [B]^ y$

Comparing Experiments 1 and 2, With [B] constant, as [A] is tripled, rate is tripled $\rightarrow$ the rate is first order with respect to [A] $\rightarrow$ x = 1

Compare experiments 2 and 3: With [A] and [B] both doubling rate increases by ~4 times $\rightarrow$ Factoring the effect of [A], rate is first order with respect to [B] $\rightarrow $ y = 1

$\Rightarrow$ Rate $ = k[A][B]$

To calculate $k$, Insert data from any experiment into the rate equation:

$\Rightarrow 5.04 \times 10^{-2} = k \times 3 \times 3 \rightarrow k = 0.0056 = 5.6 \times 10^{-3} M^{-1}s^{-1}$

Ask Question

Tag:MathPhyChemBioOther

Take Test

...