Question

A particular uncatalyzed reaction proceeds 200 times faster at 45 degrees celcius than at 0 degrees celsius

A. Calculate the activation energy for the reaction

B. when the reaction at 45 degrees celsius is catalyzed, the reaction rate increases by a factor or 500. calculate the activation energy for the catalyzed reaction?

Answer 1

Solution

A) given data

Temperature T1 = 0 C +273 = 273 K , rate = 1

Temperature T2 = 45 C +273 = 318 K   rate = 200

Ea = ?

we can use the Arhhenius equation to calculate the activation energy of the reaction

ln[[rate 2]/[rate1] ] = Ea /R [(1/T1)-(1/T2)]

R= 8.314 J per mol K

lets put the values in the formula and calculate the Ea

ln[200/1] = Ea/ 8.314 J per mol K [(1/273)-(1/318)]

5.30 = Ea/8.314 J per mol K * 0.00051835

Ea = 5.30 * 8.314 J per mol K / 0.00051835 K

Ea = 85008 J per mol

Lets convert this value from Joule to kJ

85008 J per mol * 1 kJ/ 1000 J = 85.0 kJ per mol

Therefore Eactivation of the uncatalyzed reaction = 85.0 kJ per mol

B) Given data

rate 1 = 200

rate 2 = 500

we will use the following equation

K cat / K uncat = exp [(-Ea cat + Ea uncat)/ (R*T)]

here R= 8.314 J per mol K gas constant

T= temperature in kelvin

lets plug the values in above formula

500 / 200 = k cat / k uncat = exp [(-Ea cat + 85000 J)/(8.314 J per mol K*318 K)]

by taking ln of both sides we get

ln [500/200] = [(-Ea cat + 85000 J)/(8.314 J per mol K*318 K)]

0.91629 * 8.314 *318 J = - Ea cat + 85000 J

2422.5371 J = -Ea + 85000 J

2422.5371 - 85000 J = -Ea cat

-82577.46 J = -Ea

multiplying thorugh out by -1 we get

82577.46 J = Ea

now lets convert it to the kJ

82577.46 J * 1 kJ/ 1000 J = 82.577 kJ