For the decomposition of HI, k is 3.52 x 10^-7 L/mol/s at 555 K and 3.95 x 10^-2 L/mol/s at 781 K. What is the activation energy?

1. Use the two point form Ea = -R times (ln k2 minus ln k1) divided by (1/T2 minus 1/T1). 2. The natural log of 3.52 x 10^-7 is -14.860 and the natural log of 3.95 x 10^-2 is -3.231. 3. The difference ln k2 minus ln k1 is -3.231 minus -14.860, which is 11.629. 4. The reciprocal temperatures are 1/781 K, that is 1.28 x 10^-3 per K, and 1/555 K, that is 1.80 x 10^-3 per K. 5. Their difference is 1.28 x 10^-3 minus 1.80 x 10^-3, which is -5.2 x 10^-4 per K. 6. So Ea = -8.314 J/mol/K times 11.629 divided by -5.2 x 10^-4 per K. 7. That gives 1.86 x 10^5 J/mol, or about 1.9 x 10^5 J/mol, matching the reported 185,900 J/mol. 8. Halving or doubling the log difference is the common slip and produces the 9.7 x 10^4 and 2.3 x 10^5 J/mol figures. _Source: OpenStax Chemistry (CC BY 4.0), Ch 12 "Kinetics", section 12.5 Collision Theory_