At 25∘C the reaction from Part A has a composition as shown in the table below.
Substance Pressure
(atm)
C2H2(g) 4.65
H2(g) 3.85
C2H6(g) 5.25×10−2
What is the free energy change, ΔG, in kilojoules for the reaction under these conditions?
C2H2(g) + 2H2(g) = C2H6(g)
dG° = -RTLn(Kp)
and
Kp = P(C2H6) / (P(C2H2) x P(H2)^2)
Kp = 5.25 × 10−2 / (4.65 x 3.85^2)
Kp = 7,62 x 10-4
So
dG°=-8.31 x 298 x Ln(7,62x10-4)
dG° = 17.779 kJ > 0
To calculate the free energy change (ΔG) for the reaction under the given conditions, you will need to use the equation:
ΔG = ΔG° + RT ln(Q)
Where:
ΔG is the free energy change,
ΔG° is the standard free energy change (at standard conditions),
R is the gas constant (8.314 J/(mol⋅K)),
T is the temperature in Kelvin,
ln is the natural logarithm, and
Q is the reaction quotient.
First, convert the given temperature from Celsius to Kelvin using the formula: T(K) = T(°C) + 273
T(K) = 25 + 273
T(K) = 298 K
Next, determine the reaction quotient (Q) by dividing the product of the partial pressures of the products by the product of the partial pressures of the reactants:
Q = (P(C2H6) × P(H2)) / P(C2H2)
Q = (5.25×10^(-2) × 3.85) / 4.65
Q ≈ 0.043427
Now, you need to find the standard free energy change (ΔG°) for the reaction. Unfortunately, the question doesn't provide that information. If you have the standard free energy change, you can substitute it into the equation. Otherwise, you won't be able to calculate the exact value of ΔG and can only calculate ΔG relative to standard conditions.
Once you have ΔG°, you substitute the values into the equation:
ΔG = ΔG° + RT ln(Q)
Remember to use the appropriate units for each variable (Joules, Kelvin, atm).
Note: Without the value of ΔG°, it is not possible to provide the exact ΔG value in kilojoules for the reaction under the given conditions.
To determine the free energy change (ΔG) for the reaction under the given conditions, we can use the equation:
ΔG = ΔG° + RT ln(Q)
where:
- ΔG is the free energy change
- ΔG° is the standard 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 (25°C = 298 K)
- ln(Q) is the natural logarithm of the reaction quotient (Q)
First, we need to calculate the reaction quotient (Q) using the given pressures of the substances. The reaction quotient is calculated by dividing the product of the partial pressures of the products by the product of the partial pressures of the reactants, each raised to the power of the stoichiometric coefficient:
Q = (P(C2H6) / P(C2H2)) * (P(H2))^3
where P(C2H6), P(C2H2), and P(H2) are the given pressures for each substance.
Substituting the given values:
Q = (5.25×10^(-2) / 4.65) * (3.85)^3
Next, we need to convert the Q value to a natural logarithm using the ln function.
ln(Q) = ln[(5.25×10^(-2) / 4.65) * (3.85)^3]
Finally, we substitute the values into the ΔG equation:
ΔG = ΔG° + RT ln(Q)
The value of ΔG° depends on the specific reaction. If you have the ΔG° value, you can substitute it here to calculate ΔG.
I'm at a slight disadvantage since I haven't seen part A but I think all you need to do is convert grams to moles, then use
dGrxn= (n*dGf products) - (n*dGf reactants)