In Part A, (Kp=2.7e42) we saw that G =242.1 kJ for the hydrogenation of acetylen
ID: 900965 • Letter: I
Question
In Part A, (Kp=2.7e42) we saw that G=242.1 kJ for the hydrogenation of acetylene under standard conditions (all pressures equal to 1 atm and the common reference temperature 298 K). In Part B, you will determine the G for the reaction under a given set of nonstandard conditions.
Part B
At 25 C the reaction from Part A has a composition as shown in the table below.
What is the free energy change, G, in kilojoules for the reaction under these conditions?
Substance Pressure(atm) C2H2(g) 4.25 H2(g) 4.85 C2H6(g) 1.25×102
Explanation / Answer
Apply dG totla equation
dG = dG° + RT*ln(Q)
Q = [products]/[reacttans[ ratio
dG = -242100 + 8.314*298 * ln(Q)
C2H2 + 2H2 <-> C2H6
Q = (C2H6)/(C2H2*H2^2)
Q = (1.25*10^-2)/((4.25)(4.85^2)) = 0.000125036
dG = -242100 + 8.314*298 * ln(0.000125036)
dG = -264365.71 J/mol
dG = -264.365 kJ/mol
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