Change In Entropy, Enthalpy And Gibbs Free Energy In The Synthesis Reaction Of Carbon-Monoxide And Chlorine

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Change In Entropy, Enthalpy And Gibbs Free Energy In The Synthesis Reaction Of Carbon-Monoxide And Chlorine

Given the information in figure 14.21, determine:

ΔS^{o}, ΔH^{o}, ΔG^{o} for the following reaction at 25^{o}C:

CO(g) + Cl_{2}(g) -------> COCl_{2}(g)

**The strings**:

S_{7}P_{5}A_{52} (Change - Chemical Change)
**The math**:

Pj Problem of Interest is of type *change* (entropy change). Entropy, enthalpy and Gibbs free energy are state variables and the problem of interest is the change in these states as a result of the systhesis reaction of CO and Cl_{2}.

Change in entropy = Entropy of products - entropy of reactancts

So, ΔS^{o} = S^{o} of COCl_{2} - (S^{o} of CO + S^{o} of Cl_{2})

So, ΔS^{o} = 69.1 -(47.3 + 53.3) = -31.5 cal/mole-^{o}K

Change in enthalpy = Enthalpy of formation of products - enthalpy of formation of reactants

So,ΔH^{o} = ΔH^{o}_{f} of COCl_{2} - (ΔH^{o}_{f} of CO + ΔH^{o}_{f} of Cl_{2})

So, ΔH^{o} = -53.3 -(-26.4 + 0) = -53.3 + 26.4 = -26.9 Kcal = -26900 cal.

Change in Gibbs free energy = ΔG^{o} = ΔH^{o} - TΔS^{o} where T is temperature in degree Kelvin.

So, ΔG^{o} = -26900 - 298(-31.5) = -17513 = -17.5 Kcal.

The values of ΔS^{o}, ΔH^{o}, ΔG^{o} are all negative. This indicates that this reaction will be spontaneous at low temperature.

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