If the rate constant of a reaction doubles when the temperature increases by 10°C, what is the approximate activation energy (Ea) of the reaction?
Practice Questions
1 question
Q1
If the rate constant of a reaction doubles when the temperature increases by 10°C, what is the approximate activation energy (Ea) of the reaction?
20 kJ/mol
40 kJ/mol
60 kJ/mol
80 kJ/mol
Using the Arrhenius equation, a doubling of the rate constant corresponds to an activation energy of approximately 40 kJ/mol for a 10°C increase.
Questions & Step-by-step Solutions
1 item
Q
Q: If the rate constant of a reaction doubles when the temperature increases by 10°C, what is the approximate activation energy (Ea) of the reaction?
Solution: Using the Arrhenius equation, a doubling of the rate constant corresponds to an activation energy of approximately 40 kJ/mol for a 10°C increase.
Steps: 6
Step 1: Understand that the rate constant (k) of a reaction increases with temperature.
Step 2: Know that the Arrhenius equation relates the rate constant to temperature and activation energy: k = A * e^(-Ea/(RT)), where A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
Step 3: Recognize that if the rate constant doubles (k2 = 2 * k1) when the temperature increases by 10°C, we can use this information to find the activation energy.
Step 4: Use the fact that a doubling of the rate constant can be approximated using the equation: ln(k2/k1) = (Ea/R) * (1/T1 - 1/T2). Since T2 is T1 + 10°C, convert this to Kelvin (10°C = 10 + 273.15 = 283.15 K).
Step 5: Simplify the equation for a small temperature change (10°C) to find that Ea is approximately 40 kJ/mol.
Step 6: Conclude that the activation energy (Ea) of the reaction is approximately 40 kJ/mol.
