JEE Chemistry 2023 Question with Solution

Given: The conversion is $\frac{1}{2}Cl_{2(g)} \rightarrow Cl^-_{(aq)}$.

Find: The enthalpy change for this conversion.

Using Hess's law, the process is written as:

$$\frac{1}{2}Cl_{2(g)} \rightarrow Cl_{(g)} \rightarrow Cl^-_{(g)} \rightarrow Cl^-_{(aq)}$$

Therefore,

$$\Delta H^\circ = \frac{1}{2} \times 240 + (-350) + (-380)$$ $$= -610$$

Therefore, the enthalpy change is $-610 \, \text{kJ mol}^{-1}$.

Given: $\Delta_{f}H^\circ = 240 \, \text{kJ mol}^{-1}$, $\Delta_{g}H^\circ = -350 \, \text{kJ mol}^{-1}$, and $\Delta_{hyd}H^\circ = -380 \, \text{kJ mol}^{-1}$.

Find: The nearest integer value of enthalpy change for $\frac{1}{2}Cl_2(g) \rightarrow Cl^-(aq)$.

The required enthalpy is obtained by adding the individual enthalpy terms:

$$\Delta H^\circ = \frac{1}{2} \times 240 + (-350) + (-380)$$

First,

$$\frac{1}{2} \times 240 = 120$$

Then,

$$120 - 350 - 380 = -610$$

The negative sign shows the process is exothermic. Hence, the required nearest integer is -610.