If ∆G˚= -190.60 kJ of the reaction
H2+Cl2 <--> 2HCl ,
what is the ∆G at -10.5˚C of the system if the pressure of H2=3.2atm and the pressure of Cl2=9.7atm and the pressure of HCl= 7.65atm ?
See your earlier post.
To determine the ∆G at -10.5°C of the system, we can use the equation:
∆G = ∆G˚ + RTln(Q)
where:
∆G is the free energy change of the system at the given conditions
∆G˚ is the standard free energy change of the reaction
R is the gas constant (8.314 J/(mol·K) or 0.008314 kJ/(mol·K))
T is the temperature in Kelvin
ln(Q) is the natural logarithm of the reaction quotient
First, let's convert the given temperature from °C to Kelvin:
T = -10.5°C + 273.15 = 262.65 K
Next, let's calculate the reaction quotient, Q:
Q = (P(HCl))^2 / (P(H2) * P(Cl2))
Given:
P(H2) = 3.2 atm
P(Cl2) = 9.7 atm
P(HCl) = 7.65 atm
Plugging in the values:
Q = (7.65)^2 / (3.2 * 9.7) = 18.212
Now, substitute the values into the equation for ∆G:
∆G = -190.60 kJ + (0.008314 kJ/(mol·K) * 262.65 K) * ln(18.212)
Calculate the natural logarithm:
ln(18.212) ≈ 2.902
∆G = -190.60 + (0.008314 * 262.65) * 2.902
∆G ≈ -190.60 + 0.686 * 2.902
∆G ≈ -190.60 + 1.990
∆G ≈ -188.61 kJ
Therefore, the ∆G at -10.5°C of the system is approximately -188.61 kJ.