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)
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15. Chemical Kinetics
Arrhenius Equation
Multiple Choice
The activation energy of a first-order reaction is 83.5 kJ/mol. The rate constant is 3.54 x 10⁻⁵ s⁻¹ at 45 °C. What is the rate constant at 65 °C?
A
1.23 x 10⁻⁴ s⁻¹
B
7.45 x 10⁻⁴ s⁻¹
C
2.89 x 10⁻⁵ s⁻¹
D
5.67 x 10⁻⁶ s⁻¹
Verified step by step guidance1
Identify the given values: activation energy (Ea) = 83.5 kJ/mol, initial rate constant (k1) = 3.54 x 10⁻⁵ s⁻¹ at T1 = 45 °C, and the temperature change to T2 = 65 °C.
Convert the temperatures from Celsius to Kelvin by adding 273.15 to each temperature. T1 = 45 + 273.15 K and T2 = 65 + 273.15 K.
Use the Arrhenius equation in its two-point form to find the new rate constant (k2): ln(k2/k1) = -(Ea/R) * (1/T2 - 1/T1), where R is the gas constant (8.314 J/mol·K).
Substitute the known values into the equation: Ea = 83.5 kJ/mol (convert to J/mol by multiplying by 1000), R = 8.314 J/mol·K, T1 and T2 in Kelvin, and k1 = 3.54 x 10⁻⁵ s⁻¹.
Solve the equation for k2 by calculating the right-hand side and then exponentiating both sides to isolate k2, which will give you the rate constant at 65 °C.
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