The activation energy of a first order reaction is 30 kJ/mol at 298K. The activation energy for the same reaction in the presence of a catalyst is 24 kJ/mol at 298K. How many times the reaction rate has changed in the presence of a catalyst?

To determine how many times the reaction rate has changed in the presence of a catalyst, we can use the Arrhenius equation, which relates the rate constant of a reaction to its activation energy. The equation is expressed as:

k = A * e^(-Ea/RT)

Where:

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

In this case, we have two scenarios: one without a catalyst and one with a catalyst. Let's denote:

• k₁ = rate constant without catalyst (Ea = 30 kJ/mol)
• k₂ = rate constant with catalyst (Ea = 24 kJ/mol)

First, we need to convert the activation energies from kJ/mol to J/mol for consistency with the gas constant:

• 30 kJ/mol = 30,000 J/mol
• 24 kJ/mol = 24,000 J/mol

Now, we can calculate the rate constants:

For the reaction without a catalyst:

k₁ = A * e^(-30000/(8.314 * 298))

For the reaction with a catalyst:

k₂ = A * e^(-24000/(8.314 * 298))

To find the ratio of the rate constants (which tells us how many times the reaction rate has changed), we can divide k₂ by k₁:

Rate change = k₂/k₁ = (A * e^(-24000/(8.314 * 298))) / (A * e^(-30000/(8.314 * 298)))

The pre-exponential factor (A) cancels out, simplifying our equation to:

Rate change = e^((-24000 + 30000)/(8.314 * 298))

Now, we can calculate the exponent:

Rate change = e^(6000/(8.314 * 298))

Calculating the value:

• 6000/(8.314 * 298) ≈ 0.241

Now, we find:

Rate change = e^(0.241) ≈ 1.272

This means that the reaction rate has increased by approximately 1.27 times in the presence of the catalyst. In practical terms, this indicates that the catalyst significantly enhances the reaction rate by lowering the activation energy required for the reaction to proceed.

Approved