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5. C(s) + CO2(g) ---> 2 CO(g)

Carbon (graphite), carbon dioxide and carbon monoxide form an equilibrium mixture as represented by the equation above.

(a) Predict the sign for the change in entropy, deltaS, for the reaction. Justify your prediction.

(b) In the table below are data that show the percent of CO in the equilibrium mixture at two different temperatures. Predict the sign for the change in enthalpy, deltaH, for the reaction. Justify your prediction.

(c) Appropriately complete the potential energy diagram for the reaction by finishing the curve on the graph below. Also clearly indicate delta H for the reaction on the graph.

(d) If the initial amount of C(s) was doubled, what would be the effect on the percent of CO in the equilibrium mixture? Justify your answer.

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The the change in enthalpy and entropy for reaction are explained. It also explains the effect on the CO in the mixture when initial amount C is doubled. The solution is detailed and well presented.

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5.
C(s) + CO2(g) ↔ 2 CO(g)

Carbon (graphite), carbon dioxide and carbon monoxide form an equilibrium mixture as represented by the equation above.

(a) Predict the sign for the change in entropy, ΔS, for the reaction. Justify your prediction.

There is 1 mole of gas in the reactants and 2 moles of gas in the product. So the product ...

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