N2+3H2→2NH3;∆rH°=-92.4kJmol-1 The Standard Enthalpy Of ...

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3. Thermodynamics
4. Recommended MCQs - 129 Questions (128 Qs)
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Select Chapter Topics:Thermodynamics' Properties and processClassification of System, Extensive & Intensive PropertiesEnthalpy & Internal energyCp & Cv First Law of ThermodynamicsThermochemistryHess's Law2nd & 3rd Law of Thermodynamics Spontaneity & EntropyGibbs Energy Change

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Q. No.Q63:

N2+3H2→2NH3;∆rH°=-92.4kJmol-1. The standard enthalpy of formation of NH3 gas in the above reaction would be:

1.	-92.4 J (mol)-1	2.	-46.2 kJ (mol)-1
3.	+46.2 J (mol)-1	4.	+92.4 kJ (mol)-1

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×Select Color:Add Note (Optional):Save changes1234Show me in NCERTView ExplanationAdd NoteMore ActionsHintsUpgrade Your PlanUpgrade now and unlock your full question bank experience with daily practice.Buy NowQ74:

The standard enthalpy of the formation of CH3OH(l) from the following data is:

\(\small{\mathrm{CH}_3 \mathrm{OH}_{(l)}+\frac{3}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g})+2 \mathrm{H}_2 \mathrm{O}_{(l)} \text {; }}\) \( \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-726 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}\)

\(\small{\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) \text {; } }\) \(\Delta_{\mathrm{c}} \mathrm{H}^{\circ}=-393 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}\)

\(\small{\mathrm{H}_{2(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{g})} \rightarrow \mathrm{H}_2 \mathrm{O}_{(l)} \text {; } } \) \(\Delta_{\mathrm{f}} \mathrm{H}^{\circ}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

1.	−239 kJ mol−1	2.	+239 kJ mol−1
3.	−47 kJ mol−1	4.	+47 kJ mol−1

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×Select Color:Add Note (Optional):Save changes1234Show me in NCERTView ExplanationAdd NoteMore ActionsPrevious DoubtsHintsUpgrade Your PlanUpgrade now and unlock your full question bank experience with daily practice.Buy NowUnlock IMPORTANT QUESTIONThis question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.✨ Perfect for quick revision & accuracy boostBuy Target BatchAccess all premium questions instantlyQ72:

∆vapH° (CCl4) = 30.5 kJ mol-1 ∆fH° (CCl4) = - 135.5 kJ mol-1 ∆aH° (C) = 715.0 kJ mol-1 ∆aH° (Cl2) = 242 kJ mol-1

The enthalpy change for the reaction

CCl4 (g) → C(g) + 4Cl (g) would be:

1.	326 kJ mol-1	2.	1304 kJ mol-1
3.	-328 kJ mol-1	4.	-1304 kJ mol-1

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×Select Color:Add Note (Optional):Save changes1234Show me in NCERTShow me in VideoView ExplanationAdd NoteMore ActionsHintsLinksUpgrade Your PlanUpgrade now and unlock your full question bank experience with daily practice.Buy Now Unlock IMPORTANT QUESTIONThis question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.✨ Perfect for quick revision & accuracy boostBuy Target BatchAccess all premium questions instantlyQ18:

For an isolated system with ∆U = 0, the ∆S value will be:

1.	Positive	2.	Negative
3.	Zero	4.	Not possible to define

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×Save changesSubtopic:  Classification of System, Extensive & Intensive Properties |Level 3: 35%-60%

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×Select Color:Add Note (Optional):Save changes1234Show me in NCERTView ExplanationAdd NoteMore ActionsPrevious DoubtsHintsUpgrade Your PlanUpgrade now and unlock your full question bank experience with daily practice.Buy NowUnlock IMPORTANT QUESTIONThis question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.✨ Perfect for quick revision & accuracy boostBuy Target BatchAccess all premium questions instantlyQ127:

The value of ∆G° for the given reaction would be: \( 2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{~g}) \rightarrow 2 \mathrm{D}(\mathrm{~g})\) (Given: ∆U° = – 10.5 kJ and ∆S° = – 44.1 J K–1)

1.	1.6 J	2.	–0.16 kJ
3.	0.16 kJ	4.	1.6 kJ

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×Select Color:Add Note (Optional):Save changes1234Show me in NCERTShow me in VideoView ExplanationAdd NoteMore ActionsPrevious DoubtsHintsLinksUpgrade Your PlanUpgrade now and unlock your full question bank experience with daily practice.Buy NowQ121:

The equilibrium constant for a reaction is 10. The value of ∆G° will be:

(R = 8.314 J K-1 mol-1 ; T = 300 K)

1. -5.74 kJ mol-1 2. - 5.74 J mol-1 3. + 4.57 kJ mol-1 4. -57.4 kJ mol-1

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×Select Color:Add Note (Optional):Save changes1234Show me in NCERTShow me in VideoView ExplanationAdd NoteMore ActionsHintsLinksUpgrade Your PlanUpgrade now and unlock your full question bank experience with daily practice.Buy Now Q78:

At standard conditions, if the change in the enthalpy for the following reaction is –109 kJ mol–1 H2(g)+Br2(g)→2HBr(g) and the bond energy of H2 and Br2 is 435 kJ mol–1 and 192 kJ mol–1 respectively, what is the bond energy (in kJ mol–1) of HBr?

1.	368	2.	736
3.	518	4.	259

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×Select Color:Add Note (Optional):Save changes1234Show me in NCERTShow me in VideoView ExplanationAdd NoteMore ActionsPrevious DoubtsHintsLinksUpgrade Your PlanUpgrade now and unlock your full question bank experience with daily practice.Buy NowQ68:

The enthalpy of formation of COg, CO2g, N2Og , and N2O4g are –110 kJ mol-1, – 393 kJ mol-1, 81 kJ mol-1, and 9.7 kJ \(\text{mol}^{- 1}\) respectively. The value of \(\left(\Delta\right)_{r} H\) for the following reaction would be:

\(\mathrm{N_{2} O_{4 \left(g\right)} + 3 CO{\left(g\right)} \rightarrow N_{2} O_{\left(g\right)} + 3CO_{2 \left(g\right)}}\)

1.	\(- 777 . 7\) \(kJ\) \(\text{mol}^{- 1}\)	2.	\(\) \(+ 777 . 7\) \(kJ\) \(\text{mol}^{- 1}\)
3.	\(\) \(+ 824 . 9\) \(kJ\) \(\text{mol}^{- 1}\)	4.	\(-\) \(345 . 4\) \(kJ\) \(\text{mol}^{- 1}\)

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The amount of heat needed to raise the temperature of 60.0 g of aluminium from 35°C to 55°C would be:

(Molar heat capacity of Al is \(24\) \(J\) \(\text{mol}^{- 1}\) \(K^{- 1}\))

1.	\(1 . 07\) \(J\)	2.	\(1 . 07\) \(kJ\)
3.	\(106 . 7\) \(kJ\)	4.	\(100 . 7\) \(kJ\)

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The reaction of cyanamide, NH2CN (s) with dioxygen, was carried out in a bomb calorimeter, and ∆U was found to be -742.7 kJ mol-1 at 298 K. \(\small{\mathrm{NH}_2 \mathrm{CN}(\mathrm{s})+\frac{3}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{N}_2(\mathrm{g})+\mathrm{CO}_2(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})}\)

The enthalpy change for the reaction at 298 K would be:

1. -741.3 kJ mol-1 2. + 753.9 kJ mol-1 3. + 772. 7 kJ mol-1 4. -845. 1 kJ mol-1

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×Select Color:Add Note (Optional):Save changes1234Show me in NCERTShow me in VideoView ExplanationAdd NoteMore ActionsPrevious DoubtsHintsLinksUpgrade Your PlanUpgrade now and unlock your full question bank experience with daily practice.Buy NowSelect Chapter Topics:Thermodynamics' Properties and processClassification of System, Extensive & Intensive PropertiesEnthalpy & Internal energyCp & Cv First Law of ThermodynamicsThermochemistryHess's Law2nd & 3rd Law of Thermodynamics Spontaneity & EntropyGibbs Energy Change

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