CH₃Cl(I)+Cl₂(g)→CH₂Cl₂+HCl(g), this reaction occurs with endothermic bond dissociation and exothermic bond formation . Here bond energy of C-H ,C-Cl,Cl-Cl and H-Cl are 430,328,243 and 433 kj/mol. What will be the value ∆H of above the reaction?

To calculate the enthalpy change (∆H) of the reaction, we need to use the bond dissociation energies of the bonds broken and formed during the reaction. The general equation to calculate the enthalpy change is:

$\Delta H = \text{Total energy of bonds broken} - \text{Total energy of bondsformed}$

Given Bond Energies:

- $E(C-H) = 430 \, \text{kJ/mol}$

- $E(C-Cl) = 328 \, \text{kJ/mol}$

- $E(Cl-Cl) = 243 \, \text{kJ/mol}$

- $E(H-Cl) = 433 \, \text{kJ/mol}$

Reaction:

$\text{CH}_3\text{Cl}(I) + \text{Cl}_2(g) \rightarrow \text{CH}_2\text{Cl}_2 +\text{HCl(g)}$

Identify the bonds broken

- 1 $C-H$ bond is broken.

- 1 $Cl-Cl$ bond is broken.

So, the total energy for bonds broken is:

$E_{\text{broken}} = E(C-H) + E(Cl-Cl) = 430 + 243 = 673 \, \text{kJ/mol}$

Identify the bonds formed

- 1 $C-Cl$ bond is formed.

- 1 $H-Cl$ bond is formed.

So, the total energy for bonds formed is:

$E_{\text{formed}} = E(C-Cl) + E(H-Cl) = 328 + 433 = 761 \, \text{kJ/mol}$

Calculate $\Delta H$

$\Delta H = E_{\text{broken}} - E_{\text{formed}} = 673 - 761 = -88 \, \text{kJ/mol}$

Thus, the enthalpy change $\Delta H$ for the reaction is $-88 \, \text{kJ/mol}$. This means the reaction is exothermic.