The decomposition of hydrogen iodide, 2HI(g) ----> 2H2(g) + I2 (g), has rate con

Question

The decomposition of hydrogen iodide, 2HI(g) ----> 2H2(g) + I2 (g), has rate constants of 9.51 x 10^-9 L/mol*s at 500 K and 1.10 x 10^-5 L/mol*s at 600 K. Find Ea.

Ea = -(8.314 J/mol*K)(ln 1.10x10^-5 L/mol*s / 9.51x10^-9 L/mol*s)((1/600)-(1/500))^-1

I keep getting answers that are around 15.5 kJ, but the book says it is 17.7 kJ. I don't understand what I am calculating wrong. Please provide step-by-step solutions that cover how each point is reached. I don't care for the final answer - just how you get there.

Thanks!

Explanation / Answer

ln(K2/K1)=E/R[(1/T1) - (1/T2)]

ln(1.10 x 10^-5 / 9.51 x 10^-9) = E/R [(1/500) - (1/600)]

7.053=E/R [3.333*10^-4]

E/R=21159.92

E=21159.92*8.314=175923.57J=175.923 KJ

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