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Chemistry Question on Chemical Kinetics

Rate constants of a reaction at 500 K and 700 K are 0.04s10.04 \, \text{s}^{-1} and 0.14s10.14 \, \text{s}^{-1}, respectively. Then, the activation energy of the reaction is:
{Given:} log3.5=0.5441,R=8.31J K1mol1\log 3.5 = 0.5441, \, R = 8.31 \, \text{J K}^{-1} \, \text{mol}^{-1}

A

182310 J

B

18500 J

C

18219 J

D

18030 J

Answer

18219 J

Explanation

Solution

We can use the Arrhenius equation to solve this problem. The two-point form of the Arrhenius equation is:
lnk2k1=EaR(1T11T2)\ln{\frac{k_2}{k_1}} = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)
where:
k1k_1 and k2k_2 are the rate constants at temperatures T1T_1 and T2T_2, respectively.
EaE_a is the **activation energy. **
R is the ideal gas constant.
Given: k1=0.04k_1 = 0.04 s1^{-1}
k2=0.14k_2 = 0.14 s1^{-1}
T1=500T_1 = 500 K
T2=700T_2 = 700 K
R = 8.31 J K1^{-1}mol1^{-1}
log3.5=0.5441\log 3.5 = 0.5441 which means ln3.5=2.303×0.5441=1.253\ln 3.5 = 2.303 \times 0.5441 = 1.253
Plugging the values into the equation:
ln0.140.04=Ea8.31(15001700)\ln{\frac{0.14}{0.04}} = \frac{E_a}{8.31} \left( \frac{1}{500} - \frac{1}{700} \right)

ln3.5=Ea8.31(700500500×700)\ln{3.5} = \frac{E_a}{8.31} \left( \frac{700 - 500}{500 \times 700} \right)

1.253=Ea8.31(200350000)1.253 = \frac{E_a}{8.31} \left( \frac{200}{350000} \right)

1.253×8.31×350000=Ea1.253 \times 8.31 \times 350000 = E_a

Ea=1.253×8.31×350000200E_a = \frac{1.253 \times 8.31 \times 350000}{200}

Ea=18219JE_a = 18219 J