How do you get Ea 121 for a and 17 for b

Consider the following two step mechanism: Step 1: C3H4O4(aq)C2H4O2(aq)+CO2(g)(slow)C2H4O2(aq)CH3COOH(aq)(fast) 16. a) ( 5pts) Given the table of data, what is the activation energy ( Ea) in kl for the process? Show your work and justify your answer. NO WORK = NO CREDTT Ea can be determined one of two ways: ln(k)=RET1+ln(A) 1. Plot lnk vs 1/T 1. m=Ea/R 2. solve for Ea 2. Use the kT2/kT1 equation, selecting two of the three data points, solve for Ea b) (5pts) How many times faster is the reaction at 333K than it is at 313K ? Show your work and justify your answer. NO WORK = NO CREDIT. Two strategies: 1. use the k333/k313=0.00717/0.0043 2. Use the kT2/kT1 equation, using the T2=333 and T1=313 and the Ea from 16a