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11 Nov 2019
2 things:
1. How did they flip 5.40=Ea/R(3.09x10^-4/K) to Ea=5.40 (K/3.09X10^-4)R
2: SOLVE 14.8!!!!
Consider the reaction between nitrogen dioxide and carbon monoxide: NO2(g) + CO(g)-â NO(g) + CO2(g) The rate constant at 701 K is measured as 2.57 M-1.s-1 and that at 895 K is measured as 567 M1-s-1. Find the activa- tion energy for the reaction in kJ/mol SORT You are given the rate constant of a reaction at two different temperatures and asked to find the activation energy T = 701 K, k,-2.57 M-1*S-1 T, = 895 K, k2-567 M-1 . s-1 GIVEN: FIND: Ea STRATEGIZE Use the two-point form of the Arrhenius equation, which relates the activation energy to the given information and R (a constant) 4(1-1) EQUATION In SOLVE Substitute the two rate constants and the two SOLUTION temperatures into the equation In .57 M-1s R 701 K 895 K 540-X 309 ; 10 .) Solve the equation for Ea, the activation energy, and convert to kJ/mol 6-5400 10 3.09 Ã 10-4 (309 x 10 )8.314 = 5.40 mol K 1.45 Ã 1051/mol = 1.5 Ã 102 kJ/mol CHECK The magnitude of the answer is reasonable. Activation energies for most reactions range from tens to hundreds of Kilojoules per mole FOR PRACTICE 14.8 Use the results from Example 14.8 and the given rate constant of the reaction at either of the two temperatures to predict the rate constant for this reaction at 525 K.
2 things:
1. How did they flip 5.40=Ea/R(3.09x10^-4/K) to Ea=5.40 (K/3.09X10^-4)R
2: SOLVE 14.8!!!!
Consider the reaction between nitrogen dioxide and carbon monoxide: NO2(g) + CO(g)-â NO(g) + CO2(g) The rate constant at 701 K is measured as 2.57 M-1.s-1 and that at 895 K is measured as 567 M1-s-1. Find the activa- tion energy for the reaction in kJ/mol SORT You are given the rate constant of a reaction at two different temperatures and asked to find the activation energy T = 701 K, k,-2.57 M-1*S-1 T, = 895 K, k2-567 M-1 . s-1 GIVEN: FIND: Ea STRATEGIZE Use the two-point form of the Arrhenius equation, which relates the activation energy to the given information and R (a constant) 4(1-1) EQUATION In SOLVE Substitute the two rate constants and the two SOLUTION temperatures into the equation In .57 M-1s R 701 K 895 K 540-X 309 ; 10 .) Solve the equation for Ea, the activation energy, and convert to kJ/mol 6-5400 10 3.09 Ã 10-4 (309 x 10 )8.314 = 5.40 mol K 1.45 Ã 1051/mol = 1.5 Ã 102 kJ/mol CHECK The magnitude of the answer is reasonable. Activation energies for most reactions range from tens to hundreds of Kilojoules per mole FOR PRACTICE 14.8 Use the results from Example 14.8 and the given rate constant of the reaction at either of the two temperatures to predict the rate constant for this reaction at 525 K.
Reid WolffLv2
23 Sep 2019