Single Choice

Consider the following reactions at 300 K: X $$\rightarrow$$ Y (uncatalysed reaction) & X $$\rightarrow$$ Y (catalysed reaction). The energy of activation is lowered by 83.14 kJmol$$^{-1}$$ for the catalysed reaction. The rate of catalysed reaction is:

A$$3\times 10^{14}$$ times that of uncatalysed reaction
Correct Answer
B$$15\times 10^{14}$$ times that of uncatalysed reaction
C25 times that of uncatalysed reaction
D22 times that of uncatalysed reaction

Solution

The Arrhenius equation for the uncatalyzed reaction is $$\mathrm{K}_{\mathrm{m}\mathrm{c}\mathrm{a}\mathrm{t}}=\mathrm{A}\mathrm{e}^{-\mathrm{E}_{\iota \mathrm{R}}/\mathrm{R}\mathrm{T}}.$$

The Arrhenius equation for the catalyzed reaction is $$\mathrm{K}_{\mathrm{c}\mathrm{a}\mathrm{t}}=\mathrm{A}\mathrm{e}^{-\mathrm{E}_{ \mathrm{R}}/\mathrm{R}\mathrm{T}}.$$

Let A' be the activation energy for the uncatalyzed reaction.

The activation energy for the catalyzed reaction will be $$(\mathrm{A'}-83.14\times 10^{3}).$$

Divide the two equations.

$$\displaystyle \frac{\mathrm{K}_{\mathrm{c}\mathrm{a}\mathrm{t}}}{\mathrm{K}_{\mathrm{m}\mathrm{c}\mathrm{a}\mathrm{t}}}=\frac{\mathrm{e}^{-(\mathrm{A'}-83.14\times 10^{3})/\mathrm{R}\mathrm{T}}}{\mathrm{e}^{-\mathrm{A'}/\mathrm{R}\mathrm{T}}}=\mathrm{e}^{83.14\times 10^{3}/\mathrm{R}\mathrm{T}}=\mathrm{e}^{33.33} \approx 3 \times 10^{14}$$

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