The rate constant for two parallel reactions were found reactions were found to be `1.0 xx 10^(–2)" dm"^(3)" mol"^(-1) s^(–1) and 3.0 xx 10^(–2)" dm"^(3)" mol"^(–1) s^(–1)`. If the corresponding energies of activation of the parallel reactions are `60.0" kJ mol"^(–1) and 70.0" kJ mol"^(–1)` respectively, what is the apparent overall energy of activation ?

Options

A. `130.0" kJ mol"^(-1)`
B. `67.5" kJ mol"^(-1)`
C. `100.0" kJ mol"^(-1)`
D. `65.0" kJ mol"^(-1)`

Answer

To find the apparent overall energy of activation for the two parallel reactions, we can use the following formula:

\[ E_{A,\text{net}} = \frac{k_1 \cdot E_1 + k_2 \cdot E_2}{k_1 + k_2} \]

Where:
- \( k_1 \) and \( k_2 \) are the rate constants for the two reactions.
- \( E_1 \) and \( E_2 \) are the activation energies for the two reactions.

### Step 1: Identify the given values
- Rate constant for the first reaction, \( k_1 = 1.0 \times 10^{-2} \, \text{dm}^3 \, \text{mol}^{-1} \, \text{s}^{-1} \)
- Rate constant for the second reaction, \( k_2 = 3.0 \times 10^{-2} \, \text{dm}^3 \, \text{mol}^{-1} \, \text{s}^{-1} \)
- Activation energy for the first reaction, \( E_1 = 60.0 \, \text{kJ/mol} \)
- Activation energy for the second reaction, \( E_2 = 70.0 \, \text{kJ/mol} \)

### Step 2: Convert activation energies to the same units
Since the activation energies are already in kJ/mol, we can proceed without conversion.

### Step 3: Calculate the numerator
Calculate \( k_1 \cdot E_1 + k_2 \cdot E_2 \):

\[ k_1 \cdot E_1 = (1.0 \times 10^{-2}) \cdot (60.0 \times 10^3) = 6.0 \times 10^2 \, \text{J/mol} \]

\[ k_2 \cdot E_2 = (3.0 \times 10^{-2}) \cdot (70.0 \times 10^3) = 21.0 \times 10^2 \, \text{J/mol} \]

Now, add these two results:

\[ k_1 \cdot E_1 + k_2 \cdot E_2 = 6.0 \times 10^2 + 21.0 \times 10^2 = 27.0 \times 10^2 \, \text{J/mol} \]

### Step 4: Calculate the denominator
Calculate \( k_1 + k_2 \):

\[ k_1 + k_2 = (1.0 \times 10^{-2}) + (3.0 \times 10^{-2}) = 4.0 \times 10^{-2} \, \text{dm}^3 \, \text{mol}^{-1} \, \text{s}^{-1} \]

### Step 5: Substitute into the formula
Now substitute the values into the formula for \( E_{A,\text{net}} \):

\[ E_{A,\text{net}} = \frac{27.0 \times 10^2}{4.0 \times 10^{-2}} \]

### Step 6: Calculate the overall activation energy
\[ E_{A,\text{net}} = \frac{27.0 \times 10^2}{4.0 \times 10^{-2}} = 6750 \, \text{J/mol} = 67.5 \, \text{kJ/mol} \]

### Final Answer
The apparent overall energy of activation is \( 67.5 \, \text{kJ/mol} \).

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