Using the following data for the reaction illustrated in Equation (18.13), determine the order of the reaction with respect to A and B, and the rate constant for the reaction:

Question

[latex]A+B\overset{k}{\rightarrow } C [/latex] (18.13)

[latex]\left[ A \right]\left( M \right)[/latex]	[latex]\left[ B \right]\left( M \right)[/latex]	Initial Rate [latex]\left( M s^{-1} \right)[/latex]
[latex]2.30 ×10^{-4}[/latex]	[latex]3.10 ×10^{-5}[/latex]	[latex]5.25 ×10^{-4}[/latex]
[latex]4.60 ×10^{-4}[/latex]	[latex]6.20 ×10^{-5}[/latex]	[latex]4.20 ×10^{-3}[/latex]
[latex]9.20 ×10^{-4}[/latex]	[latex]6.20 ×10^{-5}[/latex]	[latex]1.68 ×10^{-2}[/latex]

Accepted Answer

Using the last two entries in the table, the order of the reaction with respect to A is

[latex]ِ\ln \left( \frac{R_{2}}{R_{3}} \right)=\alpha\ln\left( \frac{\left[ A \right]_{2}}{\left[ A \right]_{3}} \right)[/latex]

[latex]\ln\left( \frac{4.20×10^{-3}}{1.68×10^{-2}} \right)=\alpha\ln\left( \frac{4.60×10^{-4}}{9.20×10^{-4}} \right)[/latex]

[latex]-1.386=\alpha\left( -0.693 \right)[/latex]

[latex]2=\alpha[/latex]

Using this result and the first two entries in the table, the order of the reaction with respect to B is given by

[latex]\frac{R_{1}}{R_{2}}=\frac{k\left[ A \right]_{1}^{2}\left[ B \right]_{1}^{\beta}}{k\left[ A \right]_{2}^{2}\left[ B \right]_{2}^{\beta}}=\frac{\left[ A \right]_{1}^{2}\left[ B \right]_{1}^{\beta}}{\left[ A \right]_{2}^{2}\left[ B\right]^{\beta}_{2}}[/latex]

[latex]\left( \frac{5.25×10^{-4}}{4.20×10^{-3}} \right)=\left( \frac{2.30×10^{-4}}{4.60×10^{-4}} \right)\left( \frac{3.10×10^{-5}}{6.20×10^{-5}} \right)^{\beta}[/latex]

[latex]0.500=\left( 0.500 \right)^{\beta}[/latex]

[latex]1=\beta[/latex]

Therefore, the reaction is second order in A, first order in B, and third order overall. Using any row from the table, the rate constant is readily determined:

[latex]R=k \left[ A \right]^{2} \left[ B \right][/latex]

[latex]5.2×10^{-4} M s^{-1}=k\left( 2.3×10^{-4} M \right)^{2}\left( 3.1 ×10^{-5} M \right)[/latex]

[latex]3.17 ×10^{8} M^{-2} s^{-1}=k[/latex]

Having determined k, the overall rate law is

[latex]R=\left( 3.1 ×10^{8} M^{-2} s^{-1} \right)\left[ A \right]^{2} \left[ B \right][/latex]

$\left[ A \right]\left( M \right)$	$\left[ B \right]\left( M \right)$	Initial Rate $\left( M s^{-1} \right)$
$2.30 ×10^{-4}$	$3.10 ×10^{-5}$	$5.25 ×10^{-4}$
$4.60 ×10^{-4}$	$6.20 ×10^{-5}$	$4.20 ×10^{-3}$
$9.20 ×10^{-4}$	$6.20 ×10^{-5}$	$1.68 ×10^{-2}$

Question 18.2: Using the following data for the reaction illustrated in Equ...

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