In a reaction, if the rate constant doubles when the temperature increases by 10
Practice Questions
Q1
In a reaction, if the rate constant doubles when the temperature increases by 10°C, what is the activation energy (Ea) approximately?
20 kJ/mol
40 kJ/mol
60 kJ/mol
80 kJ/mol
Questions & Step-by-Step Solutions
In a reaction, if the rate constant doubles when the temperature increases by 10°C, what is the activation energy (Ea) approximately?
Step 1: Understand the Arrhenius equation, which relates the rate constant (k) to temperature (T) and activation energy (Ea). The equation is k = A * e^(-Ea/(RT)), where A is the pre-exponential factor, R is the gas constant, and T is the temperature in Kelvin.
Step 2: Recognize that if the rate constant doubles (k2 = 2 * k1) when the temperature increases by 10°C, we can set up the equation: 2 * k1 = A * e^(-Ea/(R(T1 + 10))) and k1 = A * e^(-Ea/(RT1)).
Step 3: Take the natural logarithm of both sides of the equation to simplify it: ln(2) = (-Ea/R) * (1/(T1 + 10) - 1/T1).
Step 4: Rearrange the equation to solve for Ea: Ea = R * ln(2) / (1/(T1 + 10) - 1/T1).
Step 5: Use the value of the gas constant R = 8.314 J/(mol·K) and approximate the temperature (T1) in Kelvin. For simplicity, assume T1 is around 298 K (25°C).
Step 6: Calculate the change in the reciprocal of the temperatures: 1/(T1 + 10) - 1/T1 = 1/(308) - 1/(298).
Step 7: Plug the values into the equation to find Ea. After calculation, you will find that Ea is approximately 40 kJ/mol.
