JEE Chemistry 2025 Question with Solution

Given: $1 \text{ mol}$ water at $10^\circ C$ is converted to ice at $-10^\circ C$.

Find: Total enthalpy change.

Break the process into three steps shown in the solution:

1. Cooling liquid water from $10^\circ C$ to $0^\circ C$
2. Freezing at $0^\circ C$
3. Cooling ice from $0^\circ C$ to $-10^\circ C$

The enthalpy changes are:

$$\Delta H_1 = -10y$$ $$\Delta H_2 = -x$$ $$\Delta H_3 = -10z$$

So,

$$\Delta H = -x - 10y - 10z$$

Now use consistent units. Since $x$ is given in $kJ/mol$ while $y$ and $z$ are in $J \text{ mol}^{-1} K^{-1}$, convert $x$ to $J/mol$:

$$x \; kJ/mol = 1000x \; J/mol$$

Therefore,

$$\Delta H = -10y - 1000x - 10z$$ $$\Delta H = -10(100x + y + z)$$

Therefore, the correct option is B.

Given: The process occurs in three thermochemical stages.

Find: Net enthalpy change for the complete conversion.

From the solution, the steps are written as liquid water at $10^\circ C$, then freezing at $0^\circ C$, then ice at $-10^\circ C$.

For stage 1, cooling liquid water through $10 \text{ K}$:

$$\Delta H_1 = -y \times 10$$

For stage 2, freezing:

$$\Delta H_2 = -x \times 1000$$

For stage 3, cooling ice through $10 \text{ K}$:

$$\Delta H_3 = -z \times 10$$

Hence,

$$\Delta H_{net} = -10y - 1000x - 10z$$

Factorising,

$$\Delta H_{net} = -10(y + 100x + z)$$

which is the same as

$$\Delta H_{net} = -10(100x + y + z)$$

Therefore, the correct option is B.