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BITSAT 2012
Exam
BITSAT
Year
2012
Subject
Chemistry
Chapter
Thermodynamics (C)

BITSAT 2012 · Chemistry · Thermodynamics (C)

In which of the following reactions, standard entropy change (Δ S^...

In which of the following reactions, standard entropy change (ΔS∘)\left(\Delta \mathrm{S}^{\circ}\right) is positive and standard Gibb's energy change (ΔG∘)\left(\Delta \mathrm{G}^{\circ}\right) decreases sharply with increasing temperature?

Options

  1. AC\mathrm{C} (graphite) +12O2( g)→CO(g)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})
  2. BCO(g)+12O2( g)→CO2( g)\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})
  3. CMg(s)+12O2( g)→MgO(s)\mathrm{Mg}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{MgO}(\mathrm{s})
  4. D12C\frac{1}{2} \mathrm{C} (graphite) +12O2( g)→12CO2( g)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \frac{1}{2} \mathrm{CO}_{2}(\mathrm{~g})

Correct answer

A. C\mathrm{C} (graphite) +12O2( g)→CO(g)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})

Step-by-step solution

Since, in the first reaction gaseous products are forming from solid carbon hence entropy will increase i.e. ΔS=+ve\Delta \mathrm{S}=+\mathrm{ve}. C(gr.)+12O2( g)→CO(g);ΔS∘=+ve\mathrm{C}(\mathrm{gr} .)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g}) ; \Delta \mathrm{S}^{\circ}=+\mathrm{ve} Since, ΔG∘→=ΔH∘−TΔS\Delta \overrightarrow{\mathrm{G}^{\circ}}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S} hence the value of Δ\Delta G decrease on increasing temperature.
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