One mole each of A_2(g)$A_{2}(g)$ and B_2(g)$B_{2}(g)$ are taken in a 1L closed flask and allowed to establish the equilibrium at 500K.
A_2(g) + B_2(g) rightleftharpoons 2AB(g)$$A_{2}(g) + B_{2}(g) \rightleftharpoons 2AB(g)$$
The value of x (in textkJ mol^-1$\text{kJ mol}^{-1}$) is .... (Nearest integer)
(Given: log K=2.2, R=8.314text J K^-1text mol^-1$\log K=2.2, R=8.314\text{ J K}^{-1}\text{ mol}^{-1}$)
Keywords:#Thermodynamics Gibbs equation#enthalpy of formation#JEE Main 2026 Morning Q73#Thermodynamics JEE Main 2026
More Thermodynamics Previous-Year Questions — Page 5
Q31jee_main_2025_24_jan_eveningEnthalpy of Neutralization
Which of the following mixing of 1M base and 1M acid leads to the largest increase in temperature?
A. \text{30 mL HCl and 30 mL NaOH}
B. \text{30 mL } \mathrm{CH_{3}COOH} \text{ and 30 mL NaOH}
C. \text{50 mL HCl and 20 mL NaOH}
D. \text{45 mL } \mathrm{CH_{3}COOH} \text{ and 25 mL NaOH}
Solution
### Related Formula
Q = n_textreacted cdot Delta H_textneutralization$$Q = n_{\text{reacted}} \cdot \Delta H_{\text{neutralization}}$$Delta T = fracQm cdot c$$\Delta T = \frac{Q}{m \cdot c}$$
### Core Logic
The temperature rise depends directly on the total heat released (Q$Q$) normalized by the total heat capacity of the resulting mixed volume (m cdot c$m \cdot c$). Let's evaluate the millimoles of mathrmH^+$\mathrm{H}^+$ and mathrmOH^-$\mathrm{OH}^-$ that react in each mixture:
1. **Option 1:** 30text mL of 1mathrmM mathrmHCl + 30text mL of 1mathrmM mathrmNaOH$30\text{ mL of } 1\mathrm{M } \mathrm{HCl} + 30\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$textReactive millimoles = 30text mmol$\text{Reactive millimoles} = 30\text{ mmol}$. Both are strong electrolytes, releasing full neutralization energy (sim -57.3text kJ/mol$\sim -57.3\text{ kJ/mol}$).
Total volume = 60text mL$60\text{ mL}$.
2. **Option 2:** 30text mL of 1mathrmM mathrmCH_3COOH + 30text mL of 1mathrmM mathrmNaOH$30\text{ mL of } 1\mathrm{M } \mathrm{CH_3COOH} + 30\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$textReactive millimoles = 30text mmol$\text{Reactive millimoles} = 30\text{ mmol}$. 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:** 50text mL of 1mathrmM mathrmHCl + 20text mL of 1mathrmM mathrmNaOH$50\text{ mL of } 1\mathrm{M } \mathrm{HCl} + 20\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$textLimiting reagent = mathrmNaOH = 20text mmol$\text{Limiting reagent} = \mathrm{NaOH} = 20\text{ mmol}$. Only 20text mmol$20\text{ mmol}$ reacts.
Total volume = 70text mL$70\text{ mL}$.
4. **Option 4:** 45text mL of 1mathrmM mathrmCH_3COOH + 25text mL of 1mathrmM mathrmNaOH$45\text{ mL of } 1\mathrm{M } \mathrm{CH_3COOH} + 25\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$textLimiting reagent = 25text mmol$\text{Limiting reagent} = 25\text{ mmol}$ weak neutralization profile.
Comparing Option 1 and Option 3, Option 1 releases significantly more heat (30text mmol$30\text{ mmol}$ vs 20text mmol$20\text{ mmol}$) into a smaller volume (60text mL$60\text{ mL}$ vs 70text mL$70\text{ mL}$), yielding the largest increase in temperature Delta T$\Delta T$.
### Pattern Recognition
To maximize Delta T$\Delta T$, 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
Q41jee_main_2025_24_jan_eveningHess's Law of Constant Heat Summation
mathrmS(g) + frac32 O_2(g)
ightarrow SO_3(g) + 2xtext kcal$$\mathrm{S(g) + \frac{3}{2} O_2(g)
ightarrow SO_3(g) + 2x\text{ kcal}}$$mathrmSO2(mathrmg) + frac12mathrmO2(mathrmg)
ightarrow mathrmSO3(mathrmg) + ytext 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 mathrmSO_2(mathrmg)$\mathrm{SO}_2(\mathrm{g})$ is given by:
A. \frac{2x}{y}\mathrm{\ kcal}
B. y - 2x\mathrm{\ kcal}
C. 2x + y\mathrm{\ kcal}
D. x + y\mathrm{\ kcal}
Solution
### Related Formula
Using Hess's Law, the enthalpy change of a net reaction can be determined by linearly combining the steps:
Delta Htextnet = sum Delta Htextproducts - sum Delta Htextreactants$$\Delta H{\text{net}} = \sum \Delta H{\text{products}} - \sum \Delta H{\text{reactants}}$$
### Core Logic
The heat of formation of mathrmSO_2(g)$\mathrm{SO}_2(g)$ corresponds to the target thermochemical equation:
textTarget: mathrmS(g) + mathrmO2(g)
ightarrow mathrmSO2(g) quad Delta H_f = ?$$\text{Target: } \mathrm{S}(g) + \mathrm{O}{2}(g)
ightarrow \mathrm{SO}{2}(g) \quad \Delta H_f = ?$$
Let's write out the given equations along with their enthalpy changes (remembering that exothermic reactions release heat, so Delta H = -Q$\Delta H = -Q$):
1. mathrmS(g) + frac32mathrmO_2(g)
ightarrow mathrmSO_3(g) quad Delta H_1 = -2xtext kcal$\mathrm{S}(g) + \frac{3}{2}\mathrm{O}_{2}(g)
ightarrow \mathrm{SO}_{3}(g) \quad \Delta H_{1} = -2x\text{ kcal}$
2. mathrmSO_2(g) + frac12mathrmO_2(g)
ightarrow mathrmSO_3(g) quad Delta H_2 = -ytext kcal$\mathrm{SO}_{2}(g) + \frac{1}{2}\mathrm{O}_{2}(g)
ightarrow \mathrm{SO}_{3}(g) \quad \Delta H_{2} = -y\text{ kcal}$
To isolate mathrmSO_2(g)$\mathrm{SO}_{2}(g)$ on the product side, subtract Equation (2) from Equation (1):
left[mathrmS(g) + frac32mathrmO2(g)
ight] - left[mathrmSO2(g) + frac12mathrmO2(g)
ight]
ightarrow mathrmSO3(g) - mathrmSO3(g)$$\left[\mathrm{S}(g) + \frac{3}{2}\mathrm{O}{2}(g)
ight] - \left[\mathrm{SO}{2}(g) + \frac{1}{2}\mathrm{O}{2}(g)
ight]
ightarrow \mathrm{SO}{3}(g) - \mathrm{SO}{3}(g)$$mathrmS(g) + mathrmO2(g)
ightarrow mathrmSO2(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 H1 - Delta H_2 = -2x - (-y) = y - 2xtext kcal$$\Delta H_f = \Delta H{1} - \Delta H_{2} = -2x - (-y) = y - 2x\text{ kcal}$$
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
Q33jee_main_2025_24_jan_morningSpontaneity and Gibbs Energy Change
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.
A. Both Delta H$\Delta H$ and Delta S$\Delta S$ are (+ve)
B.Delta H$\Delta H$ is (-ve) but Delta S$\Delta S$ is (+ve)
C.Delta H$\Delta H$ is (+ve) but Delta S$\Delta S$ is (-ve)
D. Both Delta H$\Delta H$ and Delta S$\Delta S$ are (-ve)
Solution
### Related Formula
Delta G = Delta H - TDelta S$$\Delta G = \Delta H - T\Delta S$$
### Core Logic
An endothermic profile specifies that Delta H > 0$\Delta H > 0$.
For the system to become spontaneous (Delta G < 0$\Delta G < 0$) specifically when shifting to higher temperatures (T$T$), the temperature-dependent entropic subtraction term (-TDelta S$-T\Delta S$) must outweigh the enthalpic barrier. This transition demands a positive structural entropy step, i.e., Delta S > 0$\Delta S > 0$.
Hence, both Delta H$\Delta H$ and Delta S$\Delta S$ 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
Q47jee_main_2025_24_jan_morningGibbs Free Energy and Equilibrium Temperature
Standard entropies of mathrmX_2, mathrmY_2$\mathrm{X}_2, \mathrm{Y}_2$ and mathrmXY_5$\mathrm{XY}_5$ are 70, 50 and 110 mathrm~J mathrm~K^-1 mathrm~mol^-1$110 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ respectively. The temperature in Kelvin at which the reaction
frac 12 mathrm X _ 2 + frac 52 mathrm Y _ 2 rightarrow mathrm X Y _ 5 quad Delta mathrm H ^ circ = - 3 5 mathrm k J mathrm m o l ^ - 1$$\frac {1}{2} \mathrm {X} _ {2} + \frac {5}{2} \mathrm {Y} _ {2} \rightarrow \mathrm {X Y} _ {5} \quad \Delta \mathrm {H} ^ {\circ} = - 3 5 \mathrm {k J} \mathrm {m o l} ^ {- 1}$$
will be at equilibrium is (Nearest integer)
Numerical Answer.Answer: 700 to 700
Solution
### Related Formula
Delta S_textrxn^0 = sum S_textproducts^0 - sum S_textreactants^0 quad textand quad T = fracDelta H^0Delta S^0 quad textat equilibrium (Delta G^0 = 0text)$$\Delta S_{\text{rxn}}^{0} = \sum S_{\text{products}}^{0} - \sum S_{\text{reactants}}^{0} \quad \text{and} \quad T = \frac{\Delta H^0}{\Delta S^0} \quad \text{at equilibrium (}\Delta G^0 = 0\text{)}$$
### Core Logic
First, calculate the standard entropy change for the reaction system (Delta S_textrxn^0$$\Delta S_{\text{rxn}}^{0}$$):
Delta S_textrxn^0 = S^0(XY_5) - left[ frac12S^0(X_2) + frac52S^0(Y_2) right]$$\Delta S_{\text{rxn}}^{0} = S^{0}(XY_5) - \left[ \frac{1}{2}S^{0}(X_2) + \frac{5}{2}S^{0}(Y_2) \right]$$Delta S_textrxn^0 = 110 - left[ left(frac12 times 70right) + left(frac52 times 50right) right] = 110 - [35 + 125]$$\Delta S_{\text{rxn}}^{0} = 110 - \left[ \left(\frac{1}{2} \times 70\right) + \left(\frac{5}{2} \times 50\right) \right] = 110 - [35 + 125]$$Delta S_textrxn^0 = 110 - 160 = -50text J K^-1text mol^-1$$\Delta S_{\text{rxn}}^{0} = 110 - 160 = -50\text{ J K}^{-1}\text{ mol}^{-1}$$
At thermodynamic equilibrium, the change in Gibbs free energy drops to zero (Delta G^0 = 0$\Delta G^0 = 0$):
0 = Delta H^0 - TDelta S^0 implies T = fracDelta H^0Delta S^0$$0 = \Delta H^0 - T\Delta S^0 \implies T = \frac{\Delta H^0}{\Delta S^0}$$
Convert the enthalpy value into Joules (Delta H^0 = -35 times 10^3text J/mol$$\Delta H^0 = -35 \times 10^3\text{ J/mol}$$) and substitute the parameters:
T = frac-35000text J mol^-1-50text J K^-1text mol^-1 = 700text Kelvin$$T = \frac{-35000\text{ J mol}^{-1}}{-50\text{ J K}^{-1}\text{ mol}^{-1}} = 700\text{ Kelvin}$$
### Pattern Recognition
Ensure all variables use matching energy units (Joules vs. Kilojoules) before setting up your final division step.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Thermodynamics
Q33jee_main_2025_28_jan_eveningFirst Law of Thermodynamics and State Functions
An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path mathrmAto mathrmBto mathrmC rightarrow mathrmDrightarrow mathrmA$\mathrm{A}\to \mathrm{B}\to \mathrm{C} \rightarrow \mathrm{D}\rightarrow \mathrm{A}$ as shown in the three cases below.
The diagram displays three distinct cyclic paths (Case I, Case II, Case III) on volume vs pressure graphs.
Choose the correct option regarding Delta U$\Delta U$:
The diagram displays three distinct cyclic paths (Case I, Case II, Case III) on volume vs pressure graphs.The diagram displays three distinct cyclic paths (Case I, Case II, Case III) on volume vs pressure graphs.
### Related Formula
For any state function like Internal Energy (U$U$), the cyclic integral over a complete closed loop is identically zero:
oint dU = 0 implies Delta U_textcyclic = 0$$\oint dU = 0 \implies \Delta U_{\text{cyclic}} = 0$$
### Core Logic
Internal energy (U$U$) depends only on the initial and final states of the thermodynamic system, not on the path followed.
In all three listed cases, the ideal gas undergoes a complete cyclic path that returns to its original configuration state A$A$.
### Step 1: Final Evaluation
Since every transformation begins and ends at point A$A$:
Delta U_textCase-I = 0$$\Delta U_{\text{Case-I}} = 0$$Delta U_textCase-II = 0$$\Delta U_{\text{Case-II}} = 0$$Delta U_textCase-III = 0$$\Delta U_{\text{Case-III}} = 0$$
Therefore, Delta Utext (Case-I) = Delta Utext (Case-II) = Delta Utext (Case-III)$\Delta U\text{ (Case-I)} = \Delta U\text{ (Case-II)} = \Delta U\text{ (Case-III)}$.
### Pattern Recognition
Do not waste time calculating path areas or values if the question asks for a state function change (Delta U$\Delta U$, Delta H$\Delta H$, Delta S$\Delta S$, Delta G$\Delta G$) over a cyclic loop. The answer is instantly zero for all cases!
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
More Thermodynamics Questions — jee_main_2026_21_jan_morning
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