For a particular chemical reaction, the rate constant at 30.0 °C is 1.38 X 10-4 L mol-1 s-1, while the value at 49.0 °C is 1.21 X 10-3 L mol-1 s-1 . What is the activation energy for this reaction?

Question

For a particular chemical reaction, the rate constant at 30.0 °C is 1.38 X 10-4  L mol-1 s-1, while the value at 49.0 °C is 1.21 X 10-3 L mol-1 s-1 . What is the activation energy for this reaction?

a. 92.8 kJ 	
b. 200 kJ
c. 40.4 kJ 
d. 343 kJ
e. 56.4 kJ

ln 1.38x 10 -4 /
1.21 x 10 - 3 = Ea (1/303) - (1/322) =1.9473

8.70= 8.314 /
1.95 x 10 ^ -4

thats all I know how to do help please!

DrBob222 · Answer

You have reversed k1 and k2 and T1 and T2 and you don't have R anywhere in the equation.
k1 = 1.38E-4, 30C or 303K
k2 = 1.21E-3, 49C or 322K

ln(k2/k1) = (1/303)-(1/322)*Ea/R
ln(1.21-3/1.38E-4) = (.00330 - 0.00311)*Ea/8314
ln 8.768 = 0.000195Ea/8.314
2.171 = 0.0000234*Ea
Ea = ?J