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Question

Chemistry Question on Chemical Kinetics

For a reaction, activation energy Ea_a=0 and the rate constant at 200K is 1.6×106^6s1^{-1}. The rate constant at 400K will be- [Given that gas constant] R= 8.314 J K1^{-1} mol1^{-1}

A

3.2 ×104^4 s1^{-1}

B

1.6 × 106^6 s1^{-1}

C

1.6 × 103^3 s1^{-1}

D

3.2 × 106^6 s1^{-1}

Answer

1.6 × 106^6 s1^{-1}

Explanation

Solution

log(K2K1)=E2.303R(1T11T2)log\left(\frac{K_{2}}{K_{1}}\right) = \frac{E}{2.303 R}\left(\frac{1}{T_{1}} - \frac{1}{T_{2}}\right)
Ea=0E_{a} = 0
log(K2K1)=0log\left(\frac{K_{2}}{K_{1}}\right) = 0
K2K1=10=1\frac{K_{2}}{K_{1}} = 10^{\circ} = 1
K2=K1\Rightarrow K_{2} = K_{1}
K2=1.6×106s1K_{2 } = 1.6 \times 10^{6} s^{-1} at 400K400K