Calculate the equilibrium constant at 116 K for the thermodynamic data in the pr

Question

Calculate the equilibrium constant at 116 K for the thermodynamic data in the previous question. Notice that K_eq is dramatically larger for a larger temperature when there is a substantial positive increase in entropy. Warm-up question Using data from the Appendix, calculate delta G degree (in kJ) at 65 degree C for the reaction: N_2 O(1 atm) + H_2(1 atm) doubleheadarrow N_2(1 atm) + H_2 O(l) (Recall, all gases at 1 atm for standard conditions) Calculate delta G (in kJ) at 65 degree C for the reaction: N_2 0(0.0059 atm) + H_2(0.25 atm) doubleheadarrow N_2(42.7 atm) +H_2 O(l) Assume that the delta H^o and delta S^o of vaporization do not change significantly with temperature. Calculate the vapor pressure of CH_3 OH at 61 degree C (in atm). CH_3 OH (I) doubleheadarrow CH_3 OH(g) ... delta H^o = 38.0 kJ and delta S^o = 112.9 J/K

Explanation / Answer

As no data of reaction is available, i am not answering Q25.

The reaction,

N2O (g) +H2  (g) ------------------> H2 O (l) +N2 (g)

The reaction occurs at 338 K and 1 atm.

Gibbs free energy of water is -237 kJ/mol

Gibbs free energy of N2O  is 104.20 kJ/mol

Gibbs free energy of H2(g) and N2 (g) is 0

°G reaction = °Gproduct ° Greactants

= -237 kJ/mol-104.20 kJ/mol=-341.2 kJ/mol

Gibbs free energy at 338 K is ,

G 338 K= Gº + RTlnQ

Q is the reaction quotient as the molar concentration is unity Q=1

G 338 K= Gº

=-341.2 kJ/mol

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