The rate of a particular reaction doubles when temp changes from 27 oC to 37 oC. Calculate the energy of activation for such a reaction.

To determine the energy of activation for a reaction where the rate doubles with a temperature increase from 27 °C to 37 °C, we can use the Arrhenius equation. This equation relates the rate constant of a reaction to temperature and activation energy. The equation is expressed as:

Arrhenius Equation

The Arrhenius equation is given by:

k = A e^(-Ea/RT)

Where:

• k = rate constant
• A = pre-exponential factor (frequency factor)
• Ea = activation energy (in joules per mole)
• R = universal gas constant (8.314 J/(mol·K))
• T = temperature (in Kelvin)

Temperature Conversion

First, we need to convert the temperatures from Celsius to Kelvin:

• 27 °C = 27 + 273.15 = 300.15 K
• 37 °C = 37 + 273.15 = 310.15 K

Rate Constant Ratio

Since the rate doubles, we can express the relationship between the rate constants at these two temperatures:

k2 = 2k1

Using the Arrhenius equation for both temperatures:

k1 = A e^(-Ea/(R * T1))

k2 = A e^(-Ea/(R * T2))

Setting Up the Equation

Now, substituting the expressions for k1 and k2 into the ratio:

2k1 = A e^(-Ea/(R * T2))

Dividing both sides by k1 gives:

2 = e^(-Ea/(R * T2)) / e^(-Ea/(R * T1))

This simplifies to:

2 = e^(Ea/(R) * (1/T1 - 1/T2))

Taking the Natural Logarithm

Taking the natural logarithm of both sides results in:

ln(2) = (Ea/R) * (1/T1 - 1/T2)

Substituting Values

Now we can substitute the known values:

• R = 8.314 J/(mol·K)
• T1 = 300.15 K
• T2 = 310.15 K
• ln(2) ≈ 0.693

Plugging these into the equation gives:

0.693 = (Ea / 8.314) * (1/300.15 - 1/310.15)

Calculating the Temperature Difference

Calculating the difference in the reciprocal temperatures:

1/300.15 - 1/310.15 ≈ 0.000332

Final Calculation

Now we can rearrange the equation to solve for Ea:

0.693 = (Ea / 8.314) * 0.000332

Ea = 0.693 * 8.314 / 0.000332

Ea ≈ 17,000 J/mol or 17 kJ/mol

Summary

The activation energy for the reaction, given that the rate doubles with a temperature increase from 27 °C to 37 °C, is approximately 17 kJ/mol. This value reflects the energy barrier that must be overcome for the reaction to proceed, highlighting the relationship between temperature and reaction rates in chemical kinetics.