Spontaneity Of Chemical Reactions
NIOS · Class 12 · Chemistry
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Define a spontaneous process and give one example.
Answer
A spontaneous process is a process that occurs in a system by itself without requiring continuous external action. Once started, it continues naturally. Example: Cooling of hot water at room temperatu…
What is the relationship between a spontaneous process and its reverse?
Answer
If a process is spontaneous in one direction, its reverse process is non-spontaneous in that direction. For example, rusting of iron (4Fe + 3O₂ → 2Fe₂O₃) is spontaneous, but the reduction of Fe₂O₃ bac…
Define entropy and explain what it measures.
Answer
Entropy (S) is a thermodynamic property that measures the degree of disorder or randomness in a system. Higher entropy means greater disorder. For states of matter: Solid < Liquid < Gas (in terms of e…
Write the mathematical expression for entropy change during a reversible process.
Answer
ΔS = q_rev / T, where ΔS is the change in entropy, q_rev is the heat supplied reversibly at constant temperature, and T is the absolute temperature in Kelvin. This equation shows that entropy change d…
Explain why entropy increases when gases mix spontaneously.
Answer
When two ideal gases mix spontaneously (as when bulb I and bulb II are connected), the particles spread out to occupy a larger volume, increasing disorder. The internal energy (ΔU) and enthalpy (ΔH) d…
What is the Second Law of Thermodynamics?
Answer
The Second Law of Thermodynamics states that all spontaneous or natural processes produce an increase in the entropy of the universe. Mathematically: ΔS_universe = ΔS_system + ΔS_surroundings > 0 for …
Calculate the entropy change for the vaporization of water at 373K, given Δ_vap H = 40.8 kJ mol⁻¹.
Answer
Formula: Δ_vap S = Δ_vap H / T. Step 1: Convert enthalpy to J mol⁻¹: 40.8 kJ mol⁻¹ = 40,800 J mol⁻¹. Step 2: Substitute values: Δ_vap S = 40,800 J mol⁻¹ / 373 K = 109.4 J K⁻¹ mol⁻¹. Answer: 109.4 J K⁻…
Calculate the entropy change for melting of ice at 273K, given Δ_fus H = 6.02 kJ mol⁻¹.
Answer
Formula: Δ_fus S = Δ_fus H / T. Step 1: Convert to J mol⁻¹: 6.02 kJ mol⁻¹ = 6,020 J mol⁻¹. Step 2: Substitute: Δ_fus S = 6,020 J mol⁻¹ / 273 K = 22.05 J K⁻¹ mol⁻¹. Answer: 22.05 J K⁻¹ mol⁻¹ (approxima…
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