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Question

Chemistry Question on Chemical Kinetics

The rate coefficient (k)(k) for a particular reactions is 1.3×104M1s11.3 \times 10^{-4}\, M^{-1}\, s^{-1} at 100C100^{\circ}C, and 1.3×103M1s11.3 \times 10^{-3}\,M^{-1}s^{-1} at 150C150^{\circ}C What is the energy of activation (Ea)(E_{a}) (in kJkJ) for this reaction ? (R=R = molar gas constant =8.314JK1mol1)= 8.314 \,JK^{-1}\,mol^{-1})

A

16

B

60

C

99

D

132

Answer

60

Explanation

Solution

According to Arrhenius equation logk2k1=Ea2.303R(1T11T2)log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303\,R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) log1.3×1031.3×104=Ea2.303×8.314[13731423]log \frac{1.3\times10^{-3}}{1.3\times10^{-4}}=\frac{E_{a}}{2.303\times8.314} \left[\frac{1}{373}-\frac{1}{423}\right] 1==Ea2.303×8.314[13731423]1==\frac{E_{a}}{2.303\times8.314}\left[\frac{1}{373}-\frac{1}{423}\right] Ea=60kJ/moleE_{a}=60\,kJ/ mole