You are studying the formation of HI(g) from its gaseous elements H2(g) and I2(g). The free energy of formation of HI(g) is –10.10 kJ/mol at 500 K. When the free energy change of the process at 500 K is zero, the reaction mixture shows partial pressures of HI and I2 as 10.0 atm and 0.001 atm, respectively. What must be the partial pressure of H2 at this time?

Question

You are studying the formation of HI(g) from its gaseous elements H2(g) and I2(g). The free energy of formation of HI(g) is –10.10 kJ/mol at 500 K. When the free energy change of the process at 500 K is zero, the reaction mixture shows partial pressures of HI and I2 as 10.0 atm and 0.001 atm, respectively. What must be the partial pressure of H2 at this time?

Answer 1

I would do this.

dG = -RTlnK
Substitute and solve for K.

Then K = p^2HI/pH2*pI2
Substitute and solve for pH2

Answer 2

Okay, I did the equation and I got P H2 : 881.06 atm. That seems very high. Where did I make a mistake?

Thank you for your help.

Answer 3

To determine the partial pressure of H2 in the reaction mixture, we need to use the equation for the free energy change of a reaction, which relates the free energy change (ΔG), the equilibrium constant (K), and the gas constant (R) to the temperature (T) and the partial pressures of the reactants and products.

ΔG = -RT * ln(K)

Here, ΔG is the free energy change, R is the gas constant (8.314 J/K·mol), T is the temperature in Kelvin (500 K in this case), ln(K) is the natural logarithm of the equilibrium constant, and K is the ratio of the products' concentrations to the reactants' concentrations.

The equilibrium constant (K) can be determined using the partial pressures of the reactants and products:

K = (P_HI / P_H2 * P_I2)

Given that P_HI = 10.0 atm and P_I2 = 0.001 atm, we need to find the partial pressure of H2 (P_H2) when ΔG = 0.

First, let's calculate the equilibrium constant (K) using the given partial pressures:

K = (10.0 atm) / (P_H2 * 0.001 atm)

Now, substitute this expression for K into the equation for ΔG:

0 = - (8.314 J/K·mol) * (500 K) * ln((10.0 atm) / (P_H2 * 0.001 atm))

Next, let's simplify the equation:

0 = - (8.314 J/K·mol) * (500 K) * ln(10000 / P_H2)

Now, we can solve for P_H2. Divide both sides of the equation by (-8.314 J/K·mol) * (500 K):

0 / [(-8.314 J/K·mol) * (500 K)] = ln(10000 / P_H2)

Take the exponential of both sides to remove the natural logarithm:

exp(0 / [(-8.314 J/K·mol) * (500 K)]) = 10000 / P_H2

Simplifying further:

1 = 10000 / P_H2

Rearrange the equation and solve for P_H2:

P_H2 = 10000 atm

Therefore, the partial pressure of H2 at the time when the free energy change of the process is zero is 10000 atm.