Calculate Delta G (in kJ) at 65 degree C for the reaction N_2O (0.0059 atm) + H_

Question

Calculate Delta G (in kJ) at 65 degree C for the reaction N_2O (0.0059 atm) + H_2 (0.25 atm) doubleheadarrow N_2(42.7 atm) + H_2O(l) Assume that the Delta H degree and Delta S degree of vaporization do not change significantly with temperature. Calculate the vapor pressure of CH_3OH at 42 degree C (in atm). CH_3OH (I) doubleheadarrow CH_3OH(g)... Delta H degree = 38.0 kJ and delta S degree = 112.9 J/K (Don'-t round unil the end. Using the exponent enlarges any round-off error.) Calculate Delta G (in kJ) at 65 degree C for the reaction N_2O (0.0059 atm) + H_2 (0.25 atm) doubleheadarrow N_2(42.7 atm) + H_2O(l) Assume that the Delta H degree and Delta S degree of vaporization do not change significantly with temperature. Calculate the vapor pressure of CH_3OH at 42 degree C (in atm). CH_3OH (I) doubleheadarrow CH_3OH(g)... Delta H degree = 38.0 kJ and delta S degree = 112.9 J/K (Don'-t round unil the end. Using the exponent enlarges any round-off error.)

Explanation / Answer

N2O + H2 <----> N2 + H2O

Kp = p(N2) /{p(N2O)*p(H2)}
Kp = 42.7/{0.0059*0.25}
= 28949

T = 65 oC = (273+65) K = 338 K

use:
Kp = Kc*(RT)^delta n
28949 = Kc*(0.0821*338)^(1-2)
28949 = Kc*(0.0821*338)^(-1)
Kc = 8.0333*10^5

Now use:
delta G= -R*T*ln Kc
= -8.314*338*ln (8.0333*10^5)
= -38208 J
= -38.21 KJ
Answer: -38.21 KJ

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