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Q.

For a certain gaseous reaction a $10 ° C$ rise of temp, from $25 ° C$ to $35 ° C$ doubles the rate of reaction. What is the value of activation energy?

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a

$\frac{10}{2.303 R \times 298 \times 308}$

b

$\frac{0.693 R \times 298 \times 308}{10}$

c

$\frac{0.693 R \times 10}{290 \times 308}$

d

$\frac{2.303 \times 10}{298 \times 308 R}$

answer is D.

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Detailed Solution

$\log 2 = \frac{E_{a}}{2.303 R} [\frac{10}{298 \times 308}]$

$E_{a}$ $=$ $\frac{0.693 R \times 298 \times 308}{10}$

Hence, option D is correct.

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