dG for the formation of Hi(g) from its gaseous elements is -20.2 kj/mol at 500k. when the partial pressure of HI is 10 atm, and I2 0.001 atm, what must the partial pressure of hydrogen be at this temperature to reduce the magnitude of dG for the reaction to 0. H2(g)+I2(g)---->2HI(g) answer: 775 atm.
I would use dG = 0 = dGo + RTlnQ but that doesn't give me 775 atm.
how would you do it then?
what is your answer?
To determine the partial pressure of hydrogen (H2) at the given conditions to reduce the magnitude of the Gibbs Free Energy change (ΔG) for the reaction to 0, we can use the equation:
ΔG = ΔG° + RT ln(Q)
Where:
ΔG is the change in Gibbs Free Energy
ΔG° is the standard Gibbs Free Energy change
R is the gas constant (8.314 J/(mol⋅K) or 0.008314 kJ/(mol⋅K))
T is the temperature in Kelvin
Q is the reaction quotient
Given:
ΔG° = -20.2 kJ/mol
T = 500 K
P(HI) = 10 atm
P(I2) = 0.001 atm
The reaction quotient (Q) is given by the partial pressure ratio of the products divided by the partial pressure ratio of the reactants:
Q = (P(HI)^2) / (P(H2) x P(I2))
Since we want ΔG to be 0, we can set ΔG equal to 0 in the equation:
0 = ΔG° + RT ln(Q)
To solve for P(H2), we need to rearrange the equation:
ΔG° = -RT ln(Q)
Substituting the given values into the equation:
-20.2 kJ/mol = -(0.008314 kJ/(mol⋅K)) × (500 K) × ln[(10 atm)^2 / (P(H2) × 0.001 atm)]
Next, let's solve this equation for P(H2):
ln[(10^2) / (P(H2) × 0.001)] = (20.2 kJ/mol) / [(0.008314 kJ/(mol⋅K)) × 500 K]
Simplifying:
ln[10000 / (P(H2) × 0.001)] = 4876.5 / (4.157 kJ)
ln[10000 / (P(H2) × 0.001)] = 1173.5506
Using the exponential function to remove the natural logarithm:
10000 / (P(H2) × 0.001) = e^1173.5506
Rearranging for P(H2):
P(H2) = 10000 / (0.001 × e^1173.5506)
P(H2) ≈ 774.65 atm
Therefore, the partial pressure of hydrogen (H2) at this temperature to reduce the magnitude of ΔG for the reaction to 0 is approximately 775 atm.