At 25 °C, the equilibrium partial pressures for the following reaction were foun

Question

At 25 °C, the equilibrium partial pressures for the following reaction were found to be PA = 5.97 atm, PB = 5.54 atm, PC = 5.99 atm, and PD = 4.04 atm.

What is the standard change in Gibbs free energy of this reaction at 25 °C?

The equilibrium constant at 239 °C for the equation,

is Kp = 4.50 × 103 bar. Calculate the value of G°rxn at 239 °C. In which direction is the reaction spontaneous when PCl3, Cl2, and PCl5 are at standard conditions?

With the temperature at 239 °C and the partial pressures of the gases of
PCL3=0.261 bar, CL2= 0.726 bar, PCL5=1.65*10^-6 bar

calculate the value of Grxn. In which direction is the reaction spontaneous under these conditions?

Explanation / Answer

The equilibrium constant at 239 °C for the equation,

is Kp = 4.50 × 103 bar. Calculate the value of G°rxn at 239 °C. In which direction is the reaction spontaneous when PCl3, Cl2, and PCl5 are at standard conditions?

With the temperature at 239 °C and the partial pressures of the gases of
PCL3=0.261 bar, CL2= 0.726 bar, PCL5=1.65*10^-6 bar

calculate the value of Grxn. In which direction is the reaction spontaneous under these conditions?

PCl5 = PCl3 + Cl2

Q = P PCl3 * P Cl2/ P Cl5

= 0.261*0.726/ 1.65*10^-6

= 1.14*10^5

Given that Kp = 4.50 × 103 bar

Here Q>K, there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants.

Q =1.14*10^5

Delta Go = - RT ln Keq
Delta Go = - 8.3145 *298 ln 4.50 × 10^3
Delta Go = - 8.3145 *298*10.71

Delta Go = -26547.3 J/ mol

G rxn= G°rxn +RT ln Q

= -26547.3 J/mole + 8.31 J / mole K)(298 K) ln (1.14*10^5)

= -26547.3 J/mole +2477.721 J / mole (11.6)
= -26547.3 J/mole +28741.6 J/ mol

=+2194.26 J/ mol

= +2.19 kJ/ mol

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