Question 59

For a reaction, given below is the graph of $$\ln k$$ vs $$\frac{1}{T}$$. The activation energy for the reaction is equal to _____ cal mol$$^{-1}$$. (Given: $$R = 2$$ cal K$$^{-1}$$ mol$$^{-1}$$)

Correct Answer: 8

The slope of the graph is given by:

$$\mathrm{\frac{-E_a}{R}}$$

From the graph:

$$\mathrm{Slope = \frac{0-20}{5}}$$

$$\mathrm{Slope = -4}$$

Thus:

$$\mathrm{\frac{-E_a}{R} = -4}$$

Given:

$$\mathrm{R = 2}$$

Therefore:

$$\mathrm{E_a = 2 \times 4}$$

$$\mathrm{E_a = 8}$$

Hence, the required integer is:

$$\mathrm{8}$$

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