Given K_net = K₁K₂/K₃ when Ea₁ = 40 kJ/mol Ea₂ = 50 kJ/mol, Ea₃ = 60 kJ/mol. Calculate value of (E_a)_net in kJ/mol

The correct answer is 30.

Step-by-Step Explanation:

According to the Arrhenius equation, the rate constant $K$ for a reaction is given by:

$K=A·e^{-\frac{E_a}{RT}}$

We are given the relation for the overall rate constant:

$K_{\text{net}}=\frac{K_1·K_2}{K_3}$

Taking the natural logarithm (ln) on both sides of the relation:

$\ln \left(K_{\text{net}}\right)=\ln \left(K_1\right)+\ln \left(K_2\right)-\ln \left(K_3\right)$

Differentiating with respect to temperature ($T$), or using the Arrhenius form for each rate constant, we get the relationship between the activation energies:

${\left(E_a\right)}_{\text{net}}=E_{a_1}+E_{a_2}-E_{a_3}$

Now, substituting the given values of activation energy into the equation:
$E_{a_1}=40\text{kJ/mol}$
$E_{a_2}=50\text{kJ/mol}$
$E_{a_3}=60\text{kJ/mol}$

Calculation:

${\left(E_a\right)}_{\text{net}}=40+50-60$

${\left(E_a\right)}_{\text{net}}=90-60=30\text{kJ/mol}$

Thus, the value of ${\left(E_a\right)}_{\text{net}}$ is 30 kJ/mol.