For the reaction of $H_{2}$ with $I_{2}$, the rate constant is $2.5 \times 10^{-4} \,…

For the reaction of $H_{2}$ with $I_{2}$, the rate constant is $2.5 \times 10^{-4} \, \text{dm}^{3}\text{mol}^{-1}\text{s}^{-1}$ at $327^{\circ}C$ and $1.0 \, \text{dm}^{3}\text{mol}^{-1}\text{s}^{-1}$ at $527^{\circ}C$. The activation energy for the reaction, in $kJ\, \text{mol}^{-1}$ is: $R = 8.314 \, J\,K^{-1}\text{mol}^{-1}$
  1. 166
  2. 59
  3. 72
  4. 150

Solution

T2=527oC800k ;k1=2.5×10-4

T1=327oC600k ;k2=1

According to Arrhenius Equation;

lnk2k1=-EaR1T2-1T1

ln1042.5=EaR1600-1800

ln4000=Ea8.31412400

Ea=165.431 kJ, Ea166 kJ/mol

Asked in: JEE Main 2019 (10 Apr Shift 2)

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