using the rest of the Arrhenius analysis Ea931 Kjmol and A

using the rest of the Arrhenius analysis ( Ea=93.1 Kj/mol and A = 4.36×10^11 M.s^-1) predict the rate constant at 347 K

E MasteringChemistry: Post Lecture Homework Chapter 14-Google Chrome Secure I https://session.masteringchemistry.com/myct/itemVie Post Lecture Homework Chapter 14 For Practice 14.7- Enhanced with Feedb Constants Periodic Table Consider the following reaction 03 (g) O2 (g) + 0(g) You may want to reference (Pages 642-648) Section 14.5 while completing this problem

Solution

we have Arrhenius equation

k = A exp ( - Ea/RT) where k is rate constant , A = 4.36 x 10^ 11 M-1s-1, Ea = 93.1 KJ/mol = 93100 J/mol

T = 347 K

hence on substituting we get

k = (4.36 x 10^ 11 M-1s-1) exp ( - 93100 J/mol / 8.314 J/molK x 347K)

= 4.36 x 10^ 11 M-1s-1 exp ( -32.27)

= 0.0042 M-1s-1