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Spontaneity of Chemical reactions

Telangana Open School (TOSS) · Class 12 · Chemistry

Flashcards for Spontaneity of Chemical reactions — Telangana Open School (TOSS) Class 12 Chemistry. Quick Q&A cards covering key concepts, definitions, and formulas.

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Card 1Spontaneous Processes

What is a spontaneous process? Give one example.

Answer

A spontaneous process occurs naturally without external help once started. Example: Rusting of iron — iron reacts with oxygen and moisture to form rust (Fe₂O₃) without any continuous external input.

Card 2Spontaneous Processes

Why is the cooling of hot water a spontaneous process?

Answer

Hot water loses heat to the cooler surroundings until thermal equilibrium is reached. This happens naturally due to temperature difference, so it’s spontaneous. The reverse (water heating up by absorb

Card 3Entropy

Define entropy (S). How is it related to disorder?

Answer

Entropy (S) is a thermodynamic property that measures the degree of randomness or disorder in a system. Higher disorder means higher entropy. For example, gases have more entropy than liquids, which h

Card 4Entropy

Arrange the following in order of increasing entropy: 1 mol H₂O(s), 1 mol H₂O(l), 1 mol H₂O(g)

Answer

H₂O(s) < H₂O(l) < H₂O(g). Solid water (ice) is most ordered (lowest entropy), liquid has moderate disorder, and gas has maximum randomness (highest entropy).

Card 5Entropy Change

What is the formula for entropy change during a reversible process?

Answer

The entropy change is given by: $$ \Delta S = \frac{q_{\text{rev}}}{T} $$ where $ q_{\text{rev}} $ is the heat transferred reversibly and $ T $ is the absolute temperature in Kelvin.

Card 6Entropy Change in Phase Transitions

Calculate the entropy change when 1 mol of ice melts at 273 K. $\Delta_{\text{fus}}H = 6.02\,\mathrm{kJ\,mol^{-1}}$

Answer

Given: - $\Delta_{\text{fus}}H = 6.02\,\mathrm{kJ\,mol^{-1}} = 6020\,\mathrm{J\,mol^{-1}}$ - $T = 273\,\mathrm{K}$ $$ \Delta_{\text{fus}}S = \frac{\Delta_{\text{fus}}H}{T} = \frac{6020}{273} = 22.05\

Card 7Entropy Change in Phase Transitions

Calculate the entropy change for vaporization of water at 373 K if $\Delta_{\text{vap}}H = 40.8\,\mathrm{kJ\,mol^{-1}}$

Answer

Given: - $\Delta_{\text{vap}}H = 40.8\,\mathrm{kJ\,mol^{-1}} = 40800\,\mathrm{J\,mol^{-1}}$ - $T = 373\,\mathrm{K}$ $$ \Delta_{\text{vap}}S = \frac{\Delta_{\text{vap}}H}{T} = \frac{40800}{373} = 109.

Card 8Second Law of Thermodynamics

What does the second law of thermodynamics say about spontaneous processes?

Answer

The second law states that for any spontaneous process, the total entropy of the universe increases: $$ \Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0 $$ At equilibrium, $

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