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Multiple Choice
The activation energy Ea for a particular reaction is 50.0 kJ/mol. How much faster is the reaction at 313 K than at 310.0 K? (R = 8.314 J/mol • K)
A
1.15 times faster
B
1.10 times faster
C
1.20 times faster
D
1.05 times faster
Verified step by step guidance
1
Identify the formula to use: The Arrhenius equation, which relates the rate constant k to the temperature T and activation energy Ea, is given by: k = A * exp(-Ea / (R * T)), where A is the pre-exponential factor.
Use the ratio of rate constants to find how much faster the reaction is: k2/k1 = exp[(Ea/R) * (1/T1 - 1/T2)], where T1 and T2 are the initial and final temperatures, respectively.
Convert the activation energy from kJ/mol to J/mol: Ea = 50.0 kJ/mol * 1000 J/kJ = 50000 J/mol.
Substitute the given values into the equation: k2/k1 = exp[(50000 J/mol) / (8.314 J/mol*K) * (1/310 K - 1/313 K)].
Calculate the exponent and then the ratio k2/k1 to determine how much faster the reaction is at 313 K compared to 310 K.