JEE Chemistry 2023 Question with Solution

Given: Enthalpies of formation are $\Delta H_f\left(CCl_4(g)\right) = -105 \, \text{kJ/mol}$, $\Delta H_f\left(H_2O(l)\right) = -242 \, \text{kJ/mol}$, $\Delta H_f\left(CO_2(g)\right) = -394 \, \text{kJ/mol}$ and $\Delta H_f\left(HCl(g)\right) = -92 \, \text{kJ/mol}$.

Find: The magnitude of enthalpy change for

$$CCl_4(g) + 2H_2O(l) \rightarrow CO_2(g) + 4HCl(g)$$

Use the relation

$$\Delta H = \sum H_p - \sum H_R$$

Substituting the values shown in the solution:

$$\Delta H = \left(-394 + 4 \times -92\right) - \left(-105 + 2 \times -242\right)$$ $$= (-394 - 368) - (-105 - 484)$$ $$= -762 - (-589)$$ $$= -173 \, \text{kJ/mol}$$

The reaction enthalpy is $-173 \, \text{kJ/mol}$, so its magnitude is $173 \, \text{kJ/mol}$. Therefore, the answer is $173$.

Given: The standard enthalpies of formation are provided.

Find: The magnitude of the enthalpy change.

First calculate the sum for the products:

$$\sum H_p = (-394 + 4 \times -92) = -762 \, \text{kJ/mol}$$

Then calculate the sum for the reactants:

$$\sum H_R = (-105 + 2 \times -242) = -589 \, \text{kJ/mol}$$

Now apply

$$\Delta H = \sum H_p - \sum H_R$$

so,

$$\Delta H = (-762) - (-589) = -173 \, \text{kJ/mol}$$

Hence, the magnitude of the enthalpy change is $173 \, \text{kJ/mol}$.

The solution contains an inconsistency in one approach where the labels for products and reactants are interchanged, but the final magnitude remains $173$.