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Chapter 12: Problem 24

Would the slope of a \(\ln (k)\) versus 1\(/ T\) plot (with temperature in kelvin) for a catalyzed reaction be more or less negative than the slope of the $\ln (k)\( versus 1\)/ T$ plot for the uncatalyzed reaction? Explain. Assume both rate laws are first-order overall.

Short Answer

Expert verified
The slope of a ln(k) versus 1/T plot for a catalyzed reaction would be less negative than the slope of the ln(k) versus 1/T plot for the uncatalyzed reaction, as a catalyst reduces the activation energy of the reaction which results in a smaller absolute value of the slope according to the Arrhenius equation.

Step by step solution

01

Understand the effect of catalysts on activation energy.

A catalyst is a substance that increases the rate of a chemical reaction by providing an alternative reaction pathway with a lower activation energy. It does not get consumed in the reaction, so it can catalyze many reactions. For a catalyzed reaction, the activation energy is lower than for an uncatalyzed reaction. Therefore, \(Ea_{catalyzed} < Ea_{uncatalyzed}\).
02

Compare the slopes of ln(k) vs. 1/T plots for catalyzed and uncatalyzed reactions.

We have earlier established that the slope of ln(k) vs. 1/T is given by \(-\frac{Ea}{R}\). Because the activation energy of the catalyzed reaction is lower, then the slope of ln(k) vs. 1/T plot for a catalyzed reaction will be less negative than the slope for the uncatalyzed reaction. Hence, when comparing the two reactions: \(-\frac{Ea_{catalyzed}}{R} > -\frac{Ea_{uncatalyzed}}{R}\)
03

Conclusion

The slope of a ln(k) versus 1/T plot for a catalyzed reaction would be less negative than the slope of the ln(k) versus 1/T plot for the uncatalyzed reaction. The reason is that a catalyst reduces the activation energy of the reaction, leading to a smaller absolute value of the slope in the Arrhenius equation.

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Most popular questions from this chapter

The thiosulfate ion \(\left(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\right)\) is oxidized by iodine as follows: $$ 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q)+\mathrm{I}_{2}(a q) \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q)+2 \mathrm{I}^{-}(a q) $$ In a certain experiment, \(7.05 \times 10^{-3}\) mol/L of $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ is consumed in the first 11.0 seconds of the reaction. Calculate the rate of consumption of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) . Calculate the rate of production of iodide ion.

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The activation energy for the reaction $$ \mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g) $$ is 125 \(\mathrm{kJ} / \mathrm{mol}\) , and \(\Delta E\) for the reaction is $-216 \mathrm{kJ} / \mathrm{mol}$ . What is the activation energy for the reverse reaction $\left[\mathrm{NO}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{CO}(g)\right] ?$
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