Thermodynamics Previous Year Questions — JEE Main Chemistry
44 past-year Thermodynamics questions from JEE Main (Chemistry) — page 4.
Q28 (2025)
The hydration energies of $\text{K}^+$ and $\text{Cl}^-$ are $-\text{x}$ and $-\text{y}\text{ kJ/mol}$ respectively. If lattice energy of $\text{KCl}$ is $-\text{z}\text{ kJ/mol}$, then the heat of solution of $\text{KCl}$ is:
- $+\text{x} - \text{y} - \text{z}$
- $\text{x} + \text{y} + \text{z}$
- $\text{z} - (\text{x} + \text{y})$
- $-\text{z} - (\text{x} + \text{y})$
### Related Formula
$$\Delta H_{\text{sol}} = \text{Lattice Energy (L.E.)} + \Delta H_{\text{hyd}}(\text{Cation}) + \Delta H_{\text{hyd}}(\text{Anion}) $$
### Core Logic
According to Hess's Law, the dissolution process can be mapped as follows:
{{SOL_IMG1}}
Given parameters:
- Lattice Energy of $\text{KCl}$ breaking into gaseous ions $= -(-\text{z}) = \text{z}\text{ kJ/mol}$ (since lattice energy released on formation is given as $-\text{z}$).
- Hydration energy of $\text{K}^+ = -\text{x}\text{ kJ/mol}$
- Hydration energy of $\text{Cl}^- = -\text{y}\text{ kJ/mol}$
### Step 1: Computation
Substituting the values into the governing formulation:
$$\Delta H_{\text{sol}} = \text{z} + (-\text{x}) + (-\text{y}) $$
$$\Delta H_{\text{sol}} = \text{z} - \text{x} - \text{y} = \text{z} - (\text{x} + \text{y}) $$
### Pattern Recognition
To dissolve an ionic crystal, energy equal to the lattice energy must be supplied (endothermic step, $+\text{z}$), and hydration releases energy (exothermic steps, $-\text{x}$ and $-\text{y}$). Net heat of solution is simply the sum of these parts: $\text{z} - \text{x} - \text{y}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Thermodynamics
Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q33 (2025)
The correct statement amongst the following is:
- $\text{The term 'standard state' implies that the temperature is } 0^{\circ}\text{C}$
- $\text{The standard state of pure gas is the pure gas at a pressure of 1 \bar and temperature 273 K}$
- $\Delta{\text{f}}\text{H}{298}^{\theta}\text{ is zero for O}(g)$
- $\Delta{\text{f}}\text{H}{500}^{\theta}\text{ is zero for O}2(g)$
### Related Formula
$$\Delta{\text{f}}H^{\theta} = 0 \quad \text{for an element in its reference/most stable standard state} $$
### Core Logic
- Standard state conditions prescribe a pressure of $1\text{ \bar}$. Temperature is not fixed by definition but is explicitly specified (often reference tables use $298.15\text{ K}$).
- Oxygen naturally and stably exists as diatomic gas molecules ($\text{O}_2(g)$) at standard thresholds.
- The enthalpy of formation of an element in its reference elemental state is identically zero at any reference temperature:
$$\Delta{\text{f}}H_{500}^{\theta}[\text{O}2(g)] = 0 $$
Conversely, atomic oxygen gas ($\text{O}(g)$) is not the reference phase, so its formation enthalpy is non-zero.
### Step 1: Verification of Options
Statement (4) accurately aligns with thermodynamic core definitions, while statement (1) and (2) mistakenly conflate standard ambient reference states with STP conditions ($273.15\text{ K}, 1\text{ atm}$).
### Pattern Recognition
Standard state definitions checklist: Pressure $= 1\text{ \bar}$. Temperature is variable/assigned independently. Elements in their most stable natural form take $\Delta{\text{f}}H^{\theta} = 0$ at all thermal profiles.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Thermodynamics
Q31 (2025)
Which of the following mixing of 1M base and 1M acid leads to the largest increase in temperature?
- \text{30 mL HCl and 30 mL NaOH}
- \text{30 mL } \mathrm{CH_{3}COOH} \text{ and 30 mL NaOH}
- \text{50 mL HCl and 20 mL NaOH}
- \text{45 mL } \mathrm{CH_{3}COOH} \text{ and 25 mL NaOH}
### Related Formula
$$Q = n_{\text{reacted}} \cdot \Delta H_{\text{neutralization}}$$
$$\Delta T = \frac{Q}{m \cdot c}$$
### Core Logic
The temperature rise depends directly on the total heat released ($Q$) normalized by the total heat capacity of the resulting mixed volume ($m \cdot c$). Let's evaluate the millimoles of $\mathrm{H}^+$ and $\mathrm{OH}^-$ that react in each mixture:
1. **Option 1:** $30\text{ mL of } 1\mathrm{M } \mathrm{HCl} + 30\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$
$\text{Reactive millimoles} = 30\text{ mmol}$. Both are strong electrolytes, releasing full neutralization energy ($\sim -57.3\text{ kJ/mol}$).
Total volume = $60\text{ mL}$.
2. **Option 2:** $30\text{ mL of } 1\mathrm{M } \mathrm{CH_3COOH} + 30\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$
$\text{Reactive millimoles} = 30\text{ mmol}$. However, since acetic acid is a weak acid, part of the heat is consumed in its ionization. Thus, less total heat is evolved compared to Option 1.
3. **Option 3:** $50\text{ mL of } 1\mathrm{M } \mathrm{HCl} + 20\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$
$\text{Limiting reagent} = \mathrm{NaOH} = 20\text{ mmol}$. Only $20\text{ mmol}$ reacts.
Total volume = $70\text{ mL}$.
4. **Option 4:** $45\text{ mL of } 1\mathrm{M } \mathrm{CH_3COOH} + 25\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$
$\text{Limiting reagent} = 25\text{ mmol}$ weak neutralization profile.
Comparing Option 1 and Option 3, Option 1 releases significantly more heat ($30\text{ mmol}$ vs $20\text{ mmol}$) into a smaller volume ($60\text{ mL}$ vs $70\text{ mL}$), yielding the largest increase in temperature $\Delta T$.
### Pattern Recognition
To maximize $\Delta T$, look for the option that maximizes the amount of reacting strong acid and strong base equivalents while keeping the total solution volume as small as possible.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Class 11 Chemistry: Equilibrium
Q41 (2025)
$$\mathrm{S(g) + \frac{3}{2} O_2(g)
ightarrow SO_3(g) + 2x\text{ kcal}}$$
$$\mathrm{SO}2(\mathrm{g}) + \frac{1}{2}\mathrm{O}2(\mathrm{g})
ightarrow \mathrm{SO}3(\mathrm{g}) + y\text{ kcal}$$
The heat of formation of $\mathrm{SO}_2(\mathrm{g})$ is given by:
- \frac{2x}{y}\mathrm{\ kcal}
- y - 2x\mathrm{\ kcal}
- 2x + y\mathrm{\ kcal}
- x + y\mathrm{\ kcal}
### Related Formula
Using Hess's Law, the enthalpy change of a net reaction can be determined by linearly combining the steps:
$$\Delta H{\text{net}} = \sum \Delta H{\text{products}} - \sum \Delta H{\text{reactants}}$$
### Core Logic
The heat of formation of $\mathrm{SO}_2(g)$ corresponds to the target thermochemical equation:
$$\text{Target: } \mathrm{S}(g) + \mathrm{O}{2}(g)
ightarrow \mathrm{SO}{2}(g) \quad \Delta H_f = ?$$
Let's write out the given equations along with their enthalpy changes (remembering that exothermic reactions release heat, so $\Delta H = -Q$):
1. $\mathrm{S}(g) + \frac{3}{2}\mathrm{O}_{2}(g)
ightarrow \mathrm{SO}_{3}(g) \quad \Delta H_{1} = -2x\text{ kcal}$
2. $\mathrm{SO}_{2}(g) + \frac{1}{2}\mathrm{O}_{2}(g)
ightarrow \mathrm{SO}_{3}(g) \quad \Delta H_{2} = -y\text{ kcal}$
To isolate $\mathrm{SO}_{2}(g)$ on the product side, subtract Equation (2) from Equation (1):
$$\left[\mathrm{S}(g) + \frac{3}{2}\mathrm{O}{2}(g)
ight] - \left[\mathrm{SO}{2}(g) + \frac{1}{2}\mathrm{O}{2}(g)
ight]
ightarrow \mathrm{SO}{3}(g) - \mathrm{SO}{3}(g)$$
$$\mathrm{S}(g) + \mathrm{O}{2}(g)
ightarrow \mathrm{SO}{2}(g)$$
Now apply the same operation to the enthalpy values:
$$\Delta H_f = \Delta H{1} - \Delta H_{2} = -2x - (-y) = y - 2x\text{ kcal}$$
This matches Option (2).
### Pattern Recognition
To isolate your target species on the desired side of the equation, use Hess's Law to add or subtract the given elemental equations. Make sure to invert the sign of the enthalpy change if you reverse a reaction.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q33 (2025)
Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option.
- Both $\Delta H$ and $\Delta S$ are (+ve)
- $\Delta H$ is (-ve) but $\Delta S$ is (+ve)
- $\Delta H$ is (+ve) but $\Delta S$ is (-ve)
- Both $\Delta H$ and $\Delta S$ are (-ve)
### Related Formula
$$\Delta G = \Delta H - T\Delta S$$
### Core Logic
An endothermic profile specifies that $\Delta H > 0$.
For the system to become spontaneous ($\Delta G < 0$) specifically when shifting to higher temperatures ($T$), the temperature-dependent entropic subtraction term ($-T\Delta S$) must outweigh the enthalpic barrier. This transition demands a positive structural entropy step, i.e., $\Delta S > 0$.
Hence, both $\Delta H$ and $\Delta S$ are positive.
### Pattern Recognition
Spontaneity driven purely by elevated thermal thresholds mandates matching positive signs for enthalpy and entropy.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Thermodynamics
