Skip to main content
Super TutorResources
Chapter 8 of 8
QuestionsNotesFlashcardsSummaryFormulasQuizConceptsSyllabusPlan
Practice Quiz

Spontaneity Of Chemical Reactions

NIOS · Class 12 · Chemistry

Practice quiz for Spontaneity Of Chemical Reactions — NIOS Class 12 Chemistry. MCQs and questions with answers to test your preparation.

45 questions30 flashcards5 concepts

Interactive on Super Tutor

Studying Spontaneity Of Chemical Reactions? Get the full interactive chapter.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan — built for practice quiz and more.

Try Super Tutor — Free

1,000+ Class 12 students started this chapter today

Quick Quiz: Spontaneity Of Chemical Reactions

0/4

Tap an answer to check it instantly. No sign-up needed for these 4.

1

For the reaction 2NO(g) + O₂(g) → 2NO₂(g), ΔH = –113 kJ mol⁻¹ and ΔS = –145 J K⁻¹ mol⁻¹. At what temperature (approximately) does this reaction change from spontaneous to non-spontaneous?

2

The standard Gibbs energy change for the reaction CO(g) + 2H₂(g) → CH₃OH(l) at 298 K is –24.8 kJ mol⁻¹. What is the equilibrium constant K at 298 K? (R = 8.314 J K⁻¹ mol⁻¹)

3

For the rusting of iron: 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s), the standard molar entropies of Fe(s), O₂(g), and Fe₂O₃(s) are 27.3, 205.0, and 87.4 J K⁻¹ mol⁻¹ respectively. What is ΔrS° for this reaction?

4

The equilibrium constant for P(s) + 3/2 Cl₂(g) ⇌ PCl₃(g) is 2.00 × 10²⁴ at 500 K. What is the value of ΔrG°? (R = 8.314 J K⁻¹ mol⁻¹)

45 Questions·
multiple choice

Sample Questions

1multiple choice
1 marks

Which of the following correctly explains why the spontaneous mixing of two ideal gases is NOT driven by an energy change?

Show answer

For ideal gases, ΔU = ΔH = 0 upon mixing at constant temperature because there are no intermolecular forces, so entropy increase is the only driving force.

Step 1: By definition, ideal gases have no intermolecular forces of attraction or repulsion. Step 2: Therefore, internal energy (U) and enthalpy (H) depend only on temperature for ideal gases, not on pressure or volume. Step 3: At constant temperature, mixing causes no change in U or H, so ΔU = ΔH = 0. Step 4: The driving force is purely the increase in entropy — the system moves to a state of greater disorder (maximum mixing). Step 5: This is a classic example showing that energy alone cannot explain all spontaneous processes; entropy is equally important.

2multiple choice
1 marks

Consider the reaction: CCl₄(l) + H₂(g) → HCl(g) + CHCl₃(l) at 298 K where ΔrH = 91.35 kJ mol⁻¹ and ΔrS = 41.5 J K⁻¹ mol⁻¹. Which statement is correct?

Show answer

ΔG = +79.0 kJ mol⁻¹, so the reaction is non-spontaneous at 298 K.

Step 1: ΔG = ΔH – TΔS. Step 2: TΔS = 298 × 41.5 × 10⁻³ = 298 × 0.0415 = 12.367 kJ mol⁻¹. Step 3: ΔG = 91.35 – 12.367 = +78.98 ≈ +79.0 kJ mol⁻¹. Step 4: Since ΔG > 0, the reaction is non-spontaneous at 298 K. Step 5: Here ΔH > 0 and ΔS > 0 (Case 3): at low temperatures, the positive ΔH dominates, making the reaction non-spontaneous. It could become spontaneous at very high temperatures where TΔS > ΔH.

3multiple choice
1 marks

Which thermodynamic statement correctly describes the relationship between ΔG of the system and ΔS of the universe at constant temperature and pressure?

Show answer

ΔG(system) = –T × ΔS(total), so a negative ΔG corresponds to positive ΔS(universe).

Step 1: For a non-isolated system at constant T and P, ΔS(surroundings) = –ΔH(system)/T. Step 2: ΔS(total) = ΔS(system) + ΔS(surroundings) = ΔS(system) – ΔH(system)/T. Step 3: Multiplying both sides by –T: –TΔS(total) = ΔH(system) – TΔS(system) = ΔG(system). Step 4: Therefore, ΔG = –TΔS(total). Step 5: Since T is always positive, ΔG < 0 ↔ ΔS(total) > 0, which is the second law criterion for spontaneity. Gibbs energy beautifully combines both system and surroundings into one property of the system alone.

4multiple choice
1 marks

For the combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), the standard Gibbs energies of formation of CH₄, CO₂, and H₂O are –50.8, –394.4, and –237.2 kJ mol⁻¹ respectively. What is ΔrG°?

Show answer

–818.0 kJ mol⁻¹

Step 1: ΔrG° = ΣΔfG°(products) – ΣΔfG°(reactants). Step 2: ΔfG°(O₂) = 0 (element in standard state). Step 3: ΔrG° = [ΔfG°(CO₂) + 2×ΔfG°(H₂O)] – [ΔfG°(CH₄) + 2×ΔfG°(O₂)]. Step 4: = [–394.4 + 2×(–237.2)] – [–50.8 + 2×0] = [–394.4 – 474.4] – [–50.8] = –868.8 + 50.8 = –818.0 kJ mol⁻¹. Step 5: The large negative ΔrG° confirms that methane combustion is highly spontaneous, which aligns with its use as a fuel in everyday life (LPG).

+41 more questions available

Practice All

Frequently Asked Questions

What are the important topics in Spontaneity Of Chemical Reactions for NIOS Class 12 Chemistry?
Key topics in Spontaneity Of Chemical Reactions include Complete Mind Map: Spontaneity of Chemical Reactions, Complete Overview of Spontaneity of Chemical Reactions, Spontaneity of Chemical Reactions: Concept Overview. These are the concepts NIOS Class 12 examiners draw on most — study them first, then practise related questions.
How to score full marks in Spontaneity Of Chemical Reactions — NIOS Class 12 Chemistry?
Understand the core concepts first, then work through the 45 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.

Sources & Official References

Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.

More resources for Spontaneity Of Chemical Reactions

All 8 chapters — NIOS Class 12 Chemistry Practice Quiz

For serious students

Get the full Spontaneity Of Chemical Reactions chapter — for free.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan for NIOS Class 12 Chemistry.

Try Super Tutor Free
Try the full chapter — Free