Question

1) Consider the following reaction: CO(g) + H2O(g) + H2(g) + CO2(B) AHP (l/mol) -110.5 -241.8 -393.5 SU/K-mol) 197.9 188.7 131.0 213.6 a. Calculate the Standard Gibbs Free Energy (AG) and determine whether the reaction is spontaneous in the forward direction at 374K. (374K was chosen because it's just above the normal boiling point of water, which is the lowest temperature where the "standard" 1atm of water vapor can exist.) b. Calculate the equilibrium constant for the reaction at 374K. Is it what you would expect, given the standard Gibbs free energy at that temperature? C. At what temperature does the reaction become spontaneous if all pressures are held at latm ("standard conditions")? d. What pressure of carbon monoxide would be required to make the reaction run forward at 900K?

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Answer 1

92 (a) XH ZAHproducts - 20 Hucactants = (-393.5) - (-110.5 – 241.8) z –393.5 +352•* 3 = -41-2 kJ/mol. Din 2-41.2.X!0² J/mol. N N N 2 z = As = E AS products - EA Sreactants (131.0 +213•6) -(197.9+188-7) 344 36 - 386.6 -42 Jkt mot! Il

T = 3746 AG = SH-TA'S = -41200 Jmolt –(374K x -42 JKAmory -41200 Jmolt + 15,708 J molt TAG = -25,492 I molt Since SG has negative value, so the reaction is spontaneous in forward direction. (6) AG = RT lnk where R z gas constant = 8.314 Sktmolt T = Temperature Ke Equilibrium constant. 4°G = - RT enk for = – 2.303 RT Log K logk - S T K = antilong (zheter K = aw -(-25,492 I mot) 2-303x8.314 I mot 43741 I K = 3-6293 x 103 Yes the equilibrium constant has a high value which is expected due to highly negative AG -

(C) AG = BH - TIS To find the temperature at which reaction becomes spontaneous, we put o G=0. AH = -41200 J molto As = -42 JK-I molt 0 - 41200 Jmol! - T (-42354 mol-'). 41200 Frott = 42 $kt mott x T 41200 K = T. IT 42 > IT = 980.95 K. so the temperature at which reaction becomes spontaneous us 980 9.5 K.

(d) At 900k SG - AM – TA'S - -41200 Jmolt - (900 K x -42 JK imol) = -41200 Jmolt + 37800 3 molt TAG = -8400 Imot1 Nao Equilibrium constant ar kpl con ib ucun THO. N Ko z XI - 1x Pro Also, G = -2.303 RT logke = -2,303 RT log the AG = 2.303RT Log Pco. Pco - antilong le Pco = autileg -3400 i molt 2-303x8.314 31 molt x9006] I Pco z 0.635 atm