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For the reaction H2 + I2 -> 2HI, rate-constant data are given at two temperatures: at T = 769 K (1/T = 1.3 x 10⁻³ K⁻¹) log10 k = 2.9, and at T = 667 K (1/T = 1.5 x 10⁻³ K⁻¹) log10 k = 1.1. The activation energy (in cal/mol) is about:
- 4 x 10⁴
- 2 x 10⁴
- 8 x 10⁴
- 3 x 10⁴
Correct answer: 4 x 10⁴
Solution
The slope of log k vs 1/T is (2.9-1.1)/(1.3e-3 - 1.5e-3) = -9000 K; Ea = -2.303*R*slope ~ 4.1 x 10⁴ cal/mol.
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