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Choose the Best Answer · Q24

Q.ΔS\Delta S is expected to be maximum for the reaction

(a) Ca(s)+12O2(g)→CaO(s)Ca(s) + \tfrac{1}{2}O_2(g) \rightarrow CaO(s)
(b) C(s)+O2(g)→CO2(g)C(s) + O_2(g) \rightarrow CO_2(g)
(c) N2(g)+O2(g)→2NO(g)N_2(g) + O_2(g) \rightarrow 2NO(g)
(d) CaCO3(s)→CaO(s)+CO2(g)CaCO_3(s) \rightarrow CaO(s) + CO_2(g)
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Step 1. Entropy change tracks most strongly with the change in moles of GAS, Δn(g)\Delta n_{(g)}, across a reaction -- more gas moles produced means much greater disorder gained.

Step 2. Check each: (a) Ca(s)+12O2(g)→CaO(s)Ca(s)+\tfrac{1}{2}O_2(g)\rightarrow CaO(s), Δng=−12\Delta n_g=-\tfrac{1}{2} (gas consumed, entropy falls).

(b) C(s)+O2(g)→CO2(g)C(s)+O_2(g)\rightarrow CO_2(g), Δng=1−1=0\Delta n_g=1-1=0 (gas moles unchanged).

(c) N2(g)+O2(g)→2NO(g)N_2(g)+O_2(g)\rightarrow2NO(g), Δng=2−2=0\Delta n_g=2-2=0 (also unchanged). …

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