Thermodynamics Previous Year Questions — JEE Main Chemistry
44 past-year Thermodynamics questions from JEE Main (Chemistry) — page 2.
Q29 (2025)
Total enthalpy change for freezing of $1 \mathrm{~mol}$ of water at $10^{\circ} \mathrm{C}$ to ice at $-10^{\circ} \mathrm{C}$ is
(Given: $\Delta_{\mathrm{fus}}\mathrm{H} = \mathrm{x}$ kJ/mol, $\mathrm{C}_{\mathrm{p}}[\mathrm{H}_2\mathrm{O}(\ell)] = \mathrm{y}\text{ J mol}^{-1}\text{ K}^{-1}$, $\mathrm{C}_{\mathrm{p}}[\mathrm{H}_2\mathrm{O}(\text{s})] = \mathrm{z}\text{ J mol}^{-1}\text{ K}^{-1}$)
- $-\mathrm{x} - 10\mathrm{y} - 10\mathrm{z}$
- $-10(100\mathrm{x} + \mathrm{y} + \mathrm{z})$
- $10(100\mathrm{x} + \mathrm{y} + \mathrm{z})$
- $\mathrm{x} - 10\mathrm{y} - 10\mathrm{z}$
### Related Formula
$$\Delta H_{\text{total}} = n C_{p(\ell)} \Delta T_1 - n\Delta H_{\text{fusion}} + n C_{p(s)} \Delta T_2$$
### Core Logic
We need to compute the enthalpy change for the pathway:
$$\mathrm{H_2O(\ell, 10^{\circ}C)} \rightarrow \mathrm{H_2O(s, -10^{\circ}C)}$$
This can be divided into three consecutive steps:
1. Cool liquid water from $10^{\circ}C$ to $0^{\circ}C$:
$$\Delta H_1 = n \cdot C_p[\mathrm{H_2O(\ell)}] \cdot (0 - 10) = 1 \cdot y \cdot (-10) = -10y \text{ J}$$
2. Freeze water to ice at $0^{\circ}C$:
$$\Delta H_2 = -n \cdot \Delta_{\text{fus}}H = -1 \cdot x \text{ kJ} = -1000x \text{ J}$$
3. Cool ice from $0^{\circ}C$ to $-10^{\circ}C$:
$$\Delta H_3 = n \cdot C_p[\mathrm{H_2O(s)}] \cdot (-10 - 0) = 1 \cdot z \cdot (-10) = -10z \text{ J}$$
{{SOL_IMG1}}
Adding these three steps yields:
$$\Delta H_{\text{total}} = -10y - 1000x - 10z = -10(100x + y + z) \text{ Joule}$$
### Pattern Recognition
Freezing is an exothermic process, so all three steps (cooling water, freezing, and cooling ice) must carry a negative sign. Factoring out $-10$ cleanly yields the expression $-10(100x+y+z)$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Thermodynamics
Q596 (2025)
Resonance in an $\text{X}_2\text{Y}$ molecule is represented as follows:
$$\text{X} = \text{X} = \text{Y} \longleftrightarrow \text{X} \equiv \text{X}^+ - \text{Y}^-$$
The experimental enthalpy of formation for gaseous $\text{X}_2\text{Y}$ is given by the reaction:
$$\text{X} \equiv \text{X}(g) + \frac{1}{2}\text{Y} = \text{Y}(g) \longrightarrow \text{X}_2\text{Y}(g) \quad \Delta H_{f(\text{exp})} = 80 \text{ kJ mol}^{-1}$$
Calculate the magnitude of the resonance energy of $\text{X}_2\text{Y}$ in $\text{kJ mol}^{-1}$ (as the nearest integer value).
**Given bond energies:**
* $\text{X} \equiv \text{X} = 940 \text{ kJ mol}^{-1}$
* $\text{X} = \text{X} = 410 \text{ kJ mol}^{-1}$
* $\text{Y} = \text{Y} = 500 \text{ kJ mol}^{-1}$
* $\text{X} = \text{Y} = 602 \text{ kJ mol}^{-1}$
Valence settings: $\text{X} : 3, \text{Y} : 2$
### Related Formula
Resonance energy equation related to experimental and theoretical formation enthalpies:
$$\Delta H_{\text{R.E.}} = \Delta H_{f(\text{exp})} - \Delta H_{f(\text{Theo})}$$
Theoretical enthalpy calculation using bond energies:
$$\Delta H_{f(\text{Theo})} = \sum \text{B.E.}_{\text{reactants}} - \sum \text{B.E.}_{\text{products}}$$
### Execution
Step 1: Write the chemical equation to calculate the theoretical enthalpy of formation based on the localized structure $X=X=Y$:
$$X \equiv X(g) + \frac{1}{2}Y=Y(g) \longrightarrow X=X=Y(g)$$
Step 2: Substitute the localized bond energies into the reactant-minus-product relation:
$$\Delta H_{f(\text{Theo})} = \left[ \text{B.E.}_{X \equiv X} + \frac{1}{2}\text{B.E.}_{Y=Y} \right] - \left[ \text{B.E.}_{X=X} + \text{B.E.}_{X=Y} \right]$$
$$\Delta H_{f(\text{Theo})} = \left[ 940 + \frac{1}{2}(500) \right] - [410 + 602]$$
$$\Delta H_{f(\text{Theo})} = [940 + 250] - 1012 = 1190 - 1012 = 178 \text{ kJ mol}^{-1}$$
Step 3: Calculate the resonance energy:
$$\Delta H_{\text{R.E.}} = \Delta H_{f(\text{exp})} - \Delta H_{f(\text{Theo})} = 80 - 178 = -98 \text{ kJ mol}^{-1}$$
Taking the magnitude as requested: $|\Delta H_{\text{R.E.}}| = 98$.
### Pattern Recognition
Resonance energy is always a stabilizing factor, meaning $\Delta H_{f(\text{exp})}$ is more exothermic (or less endothermic) than the theoretical localized state. The magnitude is simply the absolute value of this difference ($|80 - 178| = 98$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Thermodynamics
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q39 (2025)
If $C(\text{diamond})
ightarrow C(\text{graphite}) + X\text{ kJ mol}^{-1}$
$$C(\text{diamond}) + O_{2}(g)
ightarrow CO_{2}(g) + Y\text{ kJ mol}^{-1}$$
$$C(\text{graphite}) + O_{2}(g)
ightarrow CO_{2}(g) + Z\text{ kJ mol}^{-1}$$
At constant temperature, then the correct relationship is:
- X = Y + Z
- -X = Y + Z
- X = -Y + Z
- X = Y - Z
### Core Logic
Let's treat the given parameters as terms for exothermic heats evolved on the product side:
1) $C(\text{diamond})
ightarrow C(\text{graphite})$, $Delta H = -X$
2) $C(\text{diamond}) + O_{2}(g)
ightarrow CO_{2}(g)$, $Delta H = -Y$
3) $C(\text{graphite}) + O_{2}(g)
ightarrow CO_{2}(g)$, $Delta H = -Z$
By subtracting Equation (3) from Equation (2):
$$C(\text{diamond}) - C(\text{graphite})
ightarrow 0 implies C(\text{diamond})
ightarrow C(\text{graphite})$$
$$Delta H = (-Y) - (-Z) = Z - Y$$
Matching this to Equation (1):
$$-X = Z - Y implies X = Y - Z$$
### Pattern Recognition
Hess's Law states that the enthalpy change of an overall reaction is equal to the sum of the enthalpy changes of its individual steps.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q34 (2025)
Ice and water are placed in a closed container at a pressure of 1 atm and temperature $273.15\mathrm{K}$ . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:
- $\text{Volume of system increases.}$
- $\text{Liquid phase disappears completely.}$
- $\text{The amount of ice decreases.}$
- $\text{The solid phase (ice) disappears completely.}$
### Core Logic
Water has a unique property where the density of the liquid phase is greater than the density of the solid phase (ice). Consequently, the molar volume of ice is larger than that of liquid water:
$$V_m(\text{ice}) > V_m(\text{water})$$
According to Le Chatelier's Principle, increasing the pressure favors the phase that occupies a smaller volume to alleviate the applied stress. Thus, shifting the system forward converts ice into liquid water:
{{SOL_IMG1}}
If the pressure is increased considerably (such as doubling it to 2 atm) at $273.15\mathrm{K}$, the melting point decreases, causing the entire solid phase (ice) to disappear completely.
### Pattern Recognition
Sees: Ice-water system under pressure change.
Trap: Assuming that an increase in pressure always favors the solid phase. Water has an anomalous phase curve with a negative slope.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Thermodynamics
Q50 (2025)
The formation enthalpies, $\Delta \mathrm{H}_{\mathrm{f}}^{\ominus}$ for $\mathrm{H}_{(\mathrm{g})}$ and $\mathrm{O}_{(\mathrm{g})}$ are 220.0 and $250.0~\mathrm{kJ~mol}^{-1}$ , respectively, at $298.15\mathrm{K}$ , and $\Delta \mathrm{H}_{\mathrm{f}}^{-}$ for $\mathrm{H}_2\mathrm{O}_{(\mathrm{g})}$ is $-242.0\mathrm{kJ\,mol}^{-1}$ at the same temperature. The average bond enthalpy of the O-H bond in water at $298.15\mathrm{K}$ is $\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ (nearest integer).$
### Related Formula
Reaction enthalpy based on atomization processes:
$$\Delta_r H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants})$$
### Step 1: Map the Dissociation Reaction
Consider the dissociation of gas phase water molecules into constituent gaseous atoms:
$$\mathrm{H}_2\mathrm{O}_{\mathrm{(g)}} \rightarrow 2\mathrm{H}_{\mathrm{(g)}} + \mathrm{O}_{\mathrm{(g)}}$$
The total energy required corresponds to breaking exactly two $\mathrm{O}-\mathrm{H}$ bonds:
$$\Delta_r H = 2 \times \text{B.E.}(\mathrm{O}-\mathrm{H})$$
### Step 2: Calculate $\Delta_r H$
Using the enthalpies of formation:
$$\Delta_r H = [2 \times \Delta_f H(\mathrm{H}_{\mathrm{(g)}}) + \Delta_f H(\mathrm{O}_{\mathrm{(g)}})] - \Delta_f H(\mathrm{H}_2\mathrm{O}_{\mathrm{(g)}})$$
$$\Delta_r H = [2 \times 220.0 + 250.0] - (-242.0)$$
$$\Delta_r H = [440.0 + 250.0] + 242.0 = 690.0 + 242.0 = 932.0\,\mathrm{kJ\,mol}^{-1}$$
### Step 3: Solve for Single Bond Enthalpy
$$2 \times \text{B.E.}(\mathrm{O}-\mathrm{H}) = 932.0$$
$$\text{B.E.}(\mathrm{O}-\mathrm{H}) = \frac{932.0}{2} = 466\,\mathrm{kJ\,mol}^{-1}$$
### Pattern Recognition
Sees: Atomization state values used to evaluate single bond metrics.
Shortcut: Remember $\text{Total Dissociation Energy} = \sum \Delta_f H(\text{atoms}) - \Delta_f H(\text{molecule})$. Halving the result gives the average bond enthalpy.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Thermodynamics
