What is the value of the activation energy?

How to find rate constant and activation energy?

• Rate constants for the reaction NO2+CO > NO+CO2 are 1.3 1/M*s at 700K and 23.0 1/M*s at 800K. 1) What is the value of the activation energy in kJ/mol? 2) What's the rate constant at 730K?

• Answer:

  1) What is the value of the activation energy in kJ/mol? Use the relation: ln (k'/k) = Ea/R ( 1/T- 1/T' ) Constants: R = 8.314 J/ K.mol (gas constant in joules) R=8.314 X 10^(-3) kJ/ K mol (gas constant in kJ) k' = 23.0 1/M s, T' = 800K k= 1.3 1/M s, T= 700K Solve for Ea ln (k'/k) = Ea/R ( 1/T- 1/T' ) (R X ln (k'/k)) / (1/T - T') = Ea Subsitute Ea=(R X ln (k'/k)) / (1/T - T') Ea=(8.314 X 10^(-3) X ln (23.0/1.3)) / (1/700-1/800) Ea= 133.768 kJ/mol 2) What's the rate constant at 730K? Step 1: We need A (preexponential factor) first. Use either Use the relation: k=Ae(-Ea/RT) A=k/(e(-Ea/RT) A=1.3/(-133.768/(8.314X 10^(-3)X700)) A=1.24801 X 10^(10) Step 2: Use the relation: k=Ae(-Ea/RT) (new temperature) k=1.24801 X 10^(10) X e(-133.768/(8.314X 10^(-3)X730)) k=3.34329 1/M.s Hope this helps, sorry if it's hard to read, yahoo doesn't really provide any formatting.