Question
Chemistry Question on Chemical Kinetics
Rate constants of a reaction at 500 K and 700 K are 0.04s−1 and 0.14s−1, respectively. Then, the activation energy of the reaction is:
{Given:} log3.5=0.5441,R=8.31J K−1mol−1
182310 J
18500 J
18219 J
18030 J
18219 J
Solution
We can use the Arrhenius equation to solve this problem. The two-point form of the Arrhenius equation is:
lnk1k2=REa(T11−T21)
where:
k1 and k2 are the rate constants at temperatures T1 and T2, respectively.
Ea is the **activation energy. **
R is the ideal gas constant.
Given: k1=0.04 s−1
k2=0.14 s−1
T1=500 K
T2=700 K
R = 8.31 J K−1mol−1
log3.5=0.5441 which means ln3.5=2.303×0.5441=1.253
Plugging the values into the equation:
ln0.040.14=8.31Ea(5001−7001)
ln3.5=8.31Ea(500×700700−500)
1.253=8.31Ea(350000200)
1.253×8.31×350000=Ea
Ea=2001.253×8.31×350000
Ea=18219J