For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction
2NO(g) + O2(g) --> 2NO2(g)
the standard change in Gibbs free energy is ΔG° = -69.0 kJ/mol. What is ΔG for this reaction at 298 K when the partial pressures are
P[NO] = 0.25 atm
P[O2] = 0.15 atm
P[NO2] = 0.8 atm
Delta G = G not + RTlnQ
==69.0 +(0.008314*298K*(ln((0.8^2)/((0.25^2)*0.15))
=-69.0 +10.46383111
=-58.53616889
dG = dGo + RTlnQ
where Q is set up the same way as Keq.
To find the change in Gibbs free energy, ΔG, for the reaction at the given partial pressures, we can use the equation:
ΔG = ΔG° + RT * ln(Q)
Where:
- ΔG is the change in Gibbs free energy
- ΔG° is the standard change in Gibbs free energy
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (298 K)
- Q is the reaction quotient, which is the ratio of the product of the concentrations (partial pressures) of the products to the product of the concentrations (partial pressures) of the reactants
In this case, the reaction quotient, Q, can be calculated by using the given partial pressures:
Q = (P[NO2]^2) / (P[NO]^2 * P[O2])
Now let's plug in the values:
ΔG = -69.0 kJ/mol + (8.314 J/(mol·K)) * (298 K) * ln((0.8^2) / ((0.25^2) * (0.15)))
First, let's convert kJ/mol to J/mol by multiplying -69.0 kJ/mol by 1000:
ΔG = -69,000 J/mol + (8.314 J/(mol·K)) * (298 K) * ln((0.8^2) / ((0.25^2) * (0.15)))
Calculating the natural logarithm of the ratio:
ΔG = -69,000 J/mol + (8.314 J/(mol·K)) * (298 K) * ln(2.56 / (0.0375))
Calculating the natural logarithm:
ΔG = -69,000 J/mol + (8.314 J/(mol·K)) * (298 K) * ln(68.267)
Using the ln value, we have:
ΔG = -69,000 J/mol + (8.314 J/(mol·K)) * (298 K) * 4.226
Multiplying the terms:
ΔG = -69,000 J/mol + 9998.908 J/mol
Summing the terms:
ΔG = -59,001.092 J/mol
Therefore, the change in Gibbs free energy, ΔG, for this reaction at 298 K and the given partial pressures is approximately -59,001.092 J/mol.
To calculate ΔG for this reaction at the given partial pressures, we can use the equation:
ΔG = ΔG° + RT ln(Q)
Where:
- ΔG is the change in Gibbs free energy for the reaction at non-standard conditions.
- ΔG° is the standard change in Gibbs free energy.
- R is the gas constant (8.314 J/(mol·K)).
- T is the temperature in Kelvin.
- ln is the natural logarithm.
- Q is the reaction quotient, which is calculated using the partial pressures of the reactants and products.
First, we need to calculate the reaction quotient (Q) using the given partial pressures:
Q = (P[NO2])^2 / (P[NO])^2 * P[O2]
Substituting the given values:
Q = (0.8 atm)^2 / (0.25 atm)^2 * 0.15 atm
Now, we plug in the values of Q, ΔG°, R, and T into the equation:
ΔG = -69.0 kJ/mol + (8.314 J/(mol·K)) * (298 K) * ln(Q)
Simplifying the equation and converting kJ to J:
ΔG = -69,000 J/mol + (8.314 J/(mol·K)) * (298 K) * ln(Q)
Now, you can calculate the value of ΔG using a calculator or software that supports natural logarithm (ln) function and perform the necessary calculations.