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The rate of reaction A doubles when the temperature is raised from 300 K to 310 K. For reaction B, whose activation energy is twice that of A, by how much must the temperature be raised from 300 K so that its rate also doubles?
- 2.45 K
- 4.92 K
- 9.84 K
- 19.67 K
Correct answer: 4.92 K
Solution
Since both reactions double their rate, (Ea/R)(1/300 - 1/T2) is the same constant; with Ea_B = 2 Ea_A, the temperature interval for B is about half that of A, giving roughly a 4.92 K rise.
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