The temperature dependence of the acid-catalyzed hydrolysis of penicillin (illustrated in Example Problem 18.7 )is investigated, and the dependence of $k_{1}$ on temperature is given in the following table. What is the activation energy and Arrhenius preexponential factor for this branch of the hydrolysis reaction?

Temperature $\left( °C \right)$ $k_{1}\left( s^{-1} \right)$

22.2 $7.0×10^{-4}$

27.2 $9.8×10^{-4}$

33.7 $1.6×10^{-3}$

38.0 $2.0×10^{-3}$

Question

The temperature dependence of the acid-catalyzed hydrolysis of penicillin (illustrated in Example Problem 18.7 )is investigated, and the dependence of $k_{1}$ on temperature is given in the following table. What is the activation energy and Arrhenius preexponential factor for this branch of the hydrolysis reaction?

Temperature $\left( °C \right)$	$k_{1}\left( s^{-1} \right)$
22.2	$7.0×10^{-4}$
27.2	$9.8×10^{-4}$
33.7	$1.6×10^{-3}$
38.0	$2.0×10^{-3}$

Temperature [latex]\left( °C \right)[/latex]	[latex]k_{1}\left( s^{-1} \right)[/latex]
22.2	[latex]7.0×10^{-4}[/latex]
27.2	[latex]9.8×10^{-4}[/latex]
33.7	[latex]1.6×10^{-3}[/latex]
38.0	[latex]2.0×10^{-3}[/latex]

Accepted Answer

A plot of [latex]\ln\left( k_{1} \right)[/latex] versus [latex]T^{-1}[/latex] is shown here:

The data are indicated by the points, and the solid line corresponds to the linear leastsquares fit to the data. The equation for the line is

[latex]\ln\left( k\right)=\left( -6300 K \right)\frac{1}{T}+14.1[/latex]

As shown in Equation (18.83), the slope of the line is equal to [latex]-E_{a}/R[/latex] such that

[latex]\ln \left( k \right)=\ln\left( A \right)-\frac{E_{a}}{R}\frac{1}{T}[/latex] (18.83)

[latex]6300 K=\frac{E_{a}}{R}\Rightarrow E_{a}=52,400 J mol^{-1}=52.4 kJ mol^{-1}[/latex]

The [latex]y[/latex] intercept is equal to [latex]\ln \left( A \right)[/latex] such that

[latex]A=e^{14.1}=1.33×16^{6}s^{-1}[/latex]

Question 18.8: The temperature dependence of the acid-catalyzed hydrolysis ...

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