In a reaction where the rate constant doubles with a 10°C increase in temperature, what is the approximate activation energy (Ea) if R = 8.314 J/(mol·K)?
Practice Questions
1 question
Q1
In a reaction where the rate constant doubles with a 10°C increase in temperature, what is the approximate activation energy (Ea) if R = 8.314 J/(mol·K)?
40 kJ/mol
80 kJ/mol
120 kJ/mol
160 kJ/mol
Using the Arrhenius equation and the temperature dependence of the rate constant, Ea can be estimated using the formula Ea = (R * ΔT * ln(2)) / (1/T1 - 1/T2). For a 10°C increase, Ea is approximately 80 kJ/mol.
Questions & Step-by-step Solutions
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Q
Q: In a reaction where the rate constant doubles with a 10°C increase in temperature, what is the approximate activation energy (Ea) if R = 8.314 J/(mol·K)?
Solution: Using the Arrhenius equation and the temperature dependence of the rate constant, Ea can be estimated using the formula Ea = (R * ΔT * ln(2)) / (1/T1 - 1/T2). For a 10°C increase, Ea is approximately 80 kJ/mol.
Steps: 7
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.
Step 2: Recognize that if the rate constant doubles (k2 = 2 * k1) with a temperature increase of 10°C, we can use this information to find the activation energy.
Step 3: Convert the temperature increase from Celsius to Kelvin. A 10°C increase is the same as a 10 K increase.
Step 4: Use the formula for activation energy derived from the Arrhenius equation: Ea = (R * ΔT * ln(2)) / (1/T1 - 1/T2). Here, ΔT is the temperature change (10 K), and ln(2) is the natural logarithm of 2.
Step 5: Calculate the values: R = 8.314 J/(mol·K), ΔT = 10 K, and ln(2) is approximately 0.693.
Step 6: Substitute the values into the formula: Ea = (8.314 J/(mol·K) * 10 K * 0.693) / (1/T1 - 1/T2). Since T1 and T2 are 10 K apart, you can simplify the denominator.
Step 7: After calculating, you will find that Ea is approximately 80 kJ/mol.
