For first order reaction, rate constant at 27°C and t°C is $1.5 \times 10^3$ and $4.5 \times 10^3$ respectively. If the activation energy of the reaction is 60 kJ, then find the temperature $t$.
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Correct Answer: 47.43
Solution and Explanation
The temperature dependence of the rate constant is given by the Arrhenius equation: \( \ln (k_2/k_1) = (E_a/R) (1/T_1 - 1/T_2) \).
Given: \( k_1 = 1.5 \times 10^3 \), \( T_1 = 300 K \), \( k_2 = 4.5 \times 10^3 \), \( E_a = 60000 J/mol \). Substituting into the formula: \( \ln(3) = (60000/8.314) (1/300 - 1/T_2) \).
Solving: \( 1.0986 = 7216.7 (0.003333 - 1/T_2) \) results in \( T_2 \approx 314.78 K \). Conversion to Celsius: \( t = 314.78 - 273 = 47.43^\circ C \).
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