The following reaction has ΔG° = -190.60 kJ: H2(g) + Cl2(g) ⇌ 2 HCl (g) All three gases are placed in a sealed vessel. What is the ΔG at -10.5 °C of the system if the pressure of H2 is 3.2 atm, the pressure of Cl2 is 9.7 atm, and the pressure of HCl is 7.65 atm?
dG = dGo + RTlnQ.
Q if you have forgotten is
(HCl)^2/(Cl2)(H2) for this reaction and plug in those concentrations in the problem. Remember R is 8.314, convert -10.5 to kelvin and remember to use dGo and dG in joules (not kJ)
To calculate the ΔG of the system at -10.5 °C, we need to use the formula:
ΔG = ΔG° + RTln(Q)
Where:
ΔG = standard free energy change
ΔG° = standard free energy change at standard conditions
R = gas constant (8.314 J/(mol·K) or 0.008314 kJ/(mol·K))
T = temperature in Kelvin
Q = reaction quotient
First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = -10.5 + 273.15
T(K) = 262.65 K
Then, we can calculate the reaction quotient (Q) using the partial pressures of the gases:
Q = (PHCl)^2 / (PH2 × PCl2)
Q = (7.65 atm)^2 / (3.2 atm × 9.7 atm)
Q = 58.03875 atm^2 / 31.04 atm^2
Q = 1.8693
Now, we can substitute the values into the equation to calculate the ΔG:
ΔG = ΔG° + RTln(Q)
ΔG = -190.60 kJ + (0.008314 kJ/(mol·K) × 262.65 K × ln(1.8693))
ΔG ≈ -190.60 kJ + (0.008314 kJ/(mol·K) × 262.65 K × 0.623)
ΔG ≈ -190.60 kJ + 137.45 kJ
ΔG ≈ -53.15 kJ
Therefore, the ΔG of the system at -10.5 °C is approximately -53.15 kJ.