For the reaction of H_2with I_2 the rate constant is 2.5 10^-4dm^3 mol^-1s^-1 at 327°C and 1.0dm^3 mol^1s^-1 at 527°C. The activation energy for the reaction, in kJ mol^-1 is : (R=8.314JK^-1 mol^-1)

Question

Accepted Answer

Correct answer: A. (a) K=e^-E_a/ RTor logK=- E_a2.303 RT So, log K_2K_1= E_a2.303R( 1T_1- 1T_2) log 12.5 10^-4= E_a8.314 2.303( 1600- 1800) 3.6= E_a8.314 2.303 200600 800 E_a=165.4 kJ/ mol ≈166 kJ/ mol

$For the reaction of H_{2} with I_{2} the rate constant is 2.5 \times 10^{- 4} d m^{3} mo l^{- 1} s^{- 1} at 327° C and 1.0 d m^{3} mo l^{1} s^{- 1} at 527° C . The activation energy for the reaction, in kJ mo l^{- 1} is : (R = 8.314 J K^{- 1} mo l^{- 1})$

Solution

More Chemical Kinetics PYQs