Calculate the change in Gibbs free energy at standard conditions (TSS) and the e

Question

Calculate the change in Gibbs free energy at standard conditions (TSS) and the equillbrium constant for each of the following CO(g) + 3 H2(g); ?" . +205.9 ki, as.. +2 14.7 J/K (a) CH4(g) + H2O(g) Go 141.9 (b) CaCO3(s) CaO(s) + CO2(g); ??., + 1 79.2 k, as., +160.2 uk 131.5k Did you use the equation relating the change in Gibbs free energy to the change in enthalpy, change in entropy, and temperature? Did you recall th relating the change in Gibbs free energy to the equilibrium constant and the t ture? Periodic Table Constants and Factors Standard State Thermodynamic Data Section 16.4 Tutorial 11. 1/G points I Answers os 1 16.4 WA.012 For the n N2O) + 3 H2(g)#2 NH30 at 25.0°C, the KC of the reaction is 5.4 x 105 and the Go is-32.7 kJ/mol. Use the given concentratior 9

Explanation / Answer

As we know that,

dG = dGo + RTln(k)

Now, at equilibrium, dG =0.

Therefore,

dGo = - RTln(k)

Now,

a.

dGo = 141.9 kJ/mol , R = 8.314 J/mol.K, T=298 K

ln(k) = -141.9*1000/8.314*298 = -57.27

Hence, k = 1.337 * 10-25

b.

a. dGo = 131. 5 kJ/mol , R = 8.314 J/mol.K, T=298 K

ln(k) = -131.5*1000/8.314*298 = -53.076

Hence, k = 8.95347 * 10-24

Yes, we must use all the relations as

dGo = dHo - TdSo

dG = dGo + RTln(k)

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