Question

The rate constants of two parallel first order reactions (I) and (II) are 1.50 $\times$ 10$^{-2}$s$^{-1}$ and 3.10 $\times$ 10$^{-2}$s$^{-1}$ respectively

If the corresponding energies of activation of the parallel reactions are Ea$_1$ = 64 kJ mol$^{-1}$ and Ea$_2$ = 80 kJ mol$^{-1}$ then find out the energy of activation of the overall reaction in kJ mol$^{-1}$.

Answer 1

74.78 kJ mol$^{-1}$

Answer 2

The problem involves two parallel first-order reactions. For such reactions, the overall rate constant ($k_{overall}$) is the sum of the individual rate constants ($k_1$ and $k_2$).

$k_{overall} = k_1 + k_2$

The Arrhenius equation relates the rate constant ($k$) to the activation energy ($E_a$) and temperature ($T$):

$k = A e^{-E_a / RT}$

where A is the pre-exponential factor and R is the gas constant.

To find the overall activation energy ($E_{a,overall}$), we can differentiate the Arrhenius equation with respect to temperature. Taking the natural logarithm of the Arrhenius equation:

$\ln k = \ln A - \frac{E_a}{RT}$

Differentiating with respect to temperature $T$:

$\frac{d(\ln k)}{dT} = 0 - \frac{E_a}{R} \left( -\frac{1}{T^2} \right) = \frac{E_a}{RT^2}$

Also, we know that $\frac{d(\ln k)}{dT} = \frac{1}{k} \frac{dk}{dT}$. So, $\frac{1}{k} \frac{dk}{dT} = \frac{E_a}{RT^2}$, which implies $\frac{dk}{dT} = k \frac{E_a}{RT^2}$.

For the overall reaction, we have:

$\frac{dk_{overall}}{dT} = \frac{dk_1}{dT} + \frac{dk_2}{dT}$

Substitute the expression for $\frac{dk}{dT}$ for each term:

$k_{overall} \frac{E_{a,overall}}{RT^2} = k_1 \frac{E_{a1}}{RT^2} + k_2 \frac{E_{a2}}{RT^2}$

Multiplying both sides by $RT^2$:

$k_{overall} E_{a,overall} = k_1 E_{a1} + k_2 E_{a2}$

Now, substitute $k_{overall} = k_1 + k_2$:

$(k_1 + k_2) E_{a,overall} = k_1 E_{a1} + k_2 E_{a2}$

Therefore, the overall activation energy is:

$E_{a,overall} = \frac{k_1 E_{a1} + k_2 E_{a2}}{k_1 + k_2}$

Given values:

$k_1 = 1.50 \times 10^{-2} \text{ s}^{-1}$ $E_{a1} = 64 \text{ kJ mol}^{-1}$ $k_2 = 3.10 \times 10^{-2} \text{ s}^{-1}$ $E_{a2} = 80 \text{ kJ mol}^{-1}$

Substitute these values into the formula:

$E_{a,overall} = \frac{(1.50 \times 10^{-2} \text{ s}^{-1})(64 \text{ kJ mol}^{-1}) + (3.10 \times 10^{-2} \text{ s}^{-1})(80 \text{ kJ mol}^{-1})}{1.50 \times 10^{-2} \text{ s}^{-1} + 3.10 \times 10^{-2} \text{ s}^{-1}}$

We can factor out $10^{-2}$ from the numerator and denominator:

$E_{a,overall} = \frac{10^{-2} (1.50 \times 64 + 3.10 \times 80)}{10^{-2} (1.50 + 3.10)}$ $E_{a,overall} = \frac{1.50 \times 64 + 3.10 \times 80}{1.50 + 3.10}$

Calculate the terms: Numerator: $1.50 \times 64 = 96.0$ $3.10 \times 80 = 248.0$ Sum of numerator terms $= 96.0 + 248.0 = 344.0$

Denominator: $1.50 + 3.10 = 4.60$

Now, calculate $E_{a,overall}$:

$E_{a,overall} = \frac{344.0}{4.60}$ $E_{a,overall} = \frac{3440}{46}$ $E_{a,overall} = \frac{1720}{23}$ $E_{a,overall} \approx 74.7826 \text{ kJ mol}^{-1}$

Rounding to two decimal places, the energy of activation of the overall reaction is $74.78 \text{ kJ mol}^{-1}$.

Explanation of the solution: For parallel first-order reactions, the overall rate constant ($k_{overall}$) is the sum of the individual rate constants ($k_1 + k_2$). The relationship between the overall activation energy ($E_{a,overall}$) and individual activation energies ($E_{a1}, E_{a2}$) is derived by differentiating the Arrhenius equation for the overall reaction with respect to temperature. This yields the formula:

$E_{a,overall} = \frac{k_1 E_{a1} + k_2 E_{a2}}{k_1 + k_2}$

Substitute the given values for $k_1$, $E_{a1}$, $k_2$, and $E_{a2}$ into this formula and compute the result.