Knowee
Questions
Features
Study Tools

For a reaction with an activation energy of 65 kJ mol¯¹, by what percentage is the rate constant decreased if the temperature is decreased from 37 °C to 22 °C?

Question

For a reaction with an activation energy of 65 kJ mol¯¹, by what percentage is the rate constant decreased if the temperature is decreased from 37 °C to 22 °C?

🧐 Not the exact question you are looking for?Go ask a question

Solution

To solve this problem, we need to use the Arrhenius equation which describes the temperature dependence of reaction rates. The equation is:

k = Ae^(-Ea/RT)

where: k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

The question asks for the percentage decrease in the rate constant (k), so we need to find the rate constants at the two temperatures and then calculate the percentage decrease.

Step 1: Convert the temperatures from Celsius to Kelvin. The formula to convert Celsius to Kelvin is K = °C + 273.15. So, 37°C = 310.15K and 22°C = 295.15K.

Step 2: Use the Arrhenius equation to find the rate constants at the two temperatures. We don't know the value of A, but we don't need it because we're looking for the ratio of the two rate constants, so A will cancel out.

At 37°C (310.15K), the rate constant k1 is: k1 = Ae^(-Ea/R(310.15))

At 22°C (295.15K), the rate constant k2 is: k2 = Ae^(-Ea/R(295.15))

Step 3: Find the ratio of the two rate constants. This is equal to the rate constant at 37°C divided by the rate constant at 22°C, or k1/k2.

k1/k2 = e^(-Ea/R(310.15)) / e^(-Ea/R(295.15))

Step 4: Simplify the equation. The right side of the equation simplifies to e^(Ea/R(1/295.15 - 1/310.15)).

Step 5: Plug in the given values for Ea and R. The activation energy Ea is given as 65 kJ mol¯¹, which is 65000 J mol¯¹. The gas constant R is 8.314 J mol¯¹K¯¹.

k1/k2 = e^(65000/8.314 * (1/295.15 - 1/310.15))

Step 6: Calculate the value of k1/k2.

Step 7: The percentage decrease in the rate constant is 100% - k1/k2 * 100%.

Please note that the above steps are for the derivation of the formula. You can directly use the formula derived in step 5 to calculate the percentage decrease in the rate constant.

This problem has been solved

Similar Questions

Rate of reaction becomes 2.5 times when temperatureis increased from 300 K to 310 K. If rate constant at300 K is K then rate constant at 310 K is

A particular chemical reaction occurs at room temperature (293 K) at half the rate that it does at 300 K. What is the activation energy for this reaction?

A certain chemical reaction has an activation energy of 0.500 × 10−19 J. What temperature is required for the reaction rate to be double what it is at 20°C?

If the rate of a reaction is 5 at 10oC then on increasing the temperature to 30oC,new rate is:

The rate constant of a reaction triples as the reaction temperature is raised by 10˚C from 25˚C to 35˚C.Calculate the activation energy for the reaction.

1/3

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.