A first-order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction.

• Given: log 2 = 0.3010
• log 4 = 0.6021
• R = 8.314 J/K mol

To determine the activation energy (Ea) of a first-order reaction, we can use the Arrhenius equation, which relates the rate constant (k) to temperature (T) and activation energy. The equation is given by:

Arrhenius Equation

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

Rate Constants at Different Temperatures

For a first-order reaction, the time taken to reach 50% completion (t1/2) is related to the rate constant (k) by:

t1/2 = 0.693/k

From this, we can express k as:

• At 300 K: k1 = 0.693/t1/2 = 0.693/40 min = 0.017325 min^-1
• At 320 K: k2 = 0.693/t1/2 = 0.693/20 min = 0.03465 min^-1

Using the Arrhenius Equation

Taking the natural logarithm of the ratio of the rate constants at the two temperatures gives:

ln(k2/k1) = -Ea/R * (1/T2 - 1/T1)

Substituting Values

We can convert the temperatures to Kelvin:

• T1 = 300 K
• T2 = 320 K

Now, substituting the values:

• k1 = 0.017325 min^-1
• k2 = 0.03465 min^-1

Calculating the left side:

ln(k2/k1) = ln(0.03465/0.017325) = ln(2) = 0.3010

Calculating Activation Energy

Now, substituting into the equation:

0.3010 = -Ea/(8.314 J/K mol) * (1/320 - 1/300)

Calculating the temperature difference:

(1/320 - 1/300) = (300 - 320)/(300 * 320) = -20/(96000) = -0.00020833 K^-1

Final Calculation

Now, substituting this back into the equation:

0.3010 = -Ea/(8.314) * (-0.00020833)

Rearranging gives:

0.3010 = Ea/(8.314 * 0.00020833)

Thus, Ea = 0.3010 * 8.314 * 0.00020833

Calculating Ea yields:

Ea ≈ 4840.5 J/mol or 48.4 kJ/mol

Result

The activation energy of the reaction is approximately 48.4 kJ/mol.