Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.
M203(s)<---> 2M(s) + 1/2 O2(g)
info given for Gf(kJ/mol):
M203= -6.00
M(s)=0
O2(g)= 0
1.) what is the standard change in Gibbs energy for rxn as written in forward direction? (kJ/mol)
2.) What is the equilibrium constant (K) of this rxn, as written in forward direction at 298K?
3.) What is the equilibrium pressure of O2(g) over M(s) at 298K? (atm)
What are the units? -6.00 what?
1.
dHrxn = (n*Gf products)-(n*dGf reactants).
2.
dG = -RTlnK
3.
Kp = pO2
To answer these questions, we need to use the standard change in Gibbs energy (ΔG°) and the equation relating ΔG°, equilibrium constant (K), and temperature (T).
1.) The standard change in Gibbs energy for the reaction as written in the forward direction (ΔG°f) can be calculated using the following equation:
ΔG°f = ΣnΔGf(products) - ΣnΔGf(reactants)
Given the values of the standard Gibbs energy of formation (ΔGf) for the species involved in the reaction:
ΔGf(M203) = -6.00 kJ/mol
ΔGf(M) = 0 kJ/mol
ΔGf(O2) = 0 kJ/mol
Substituting the values into the equation, we get:
ΔG°f = (2 × ΔGf(M)) + (1/2 × ΔGf(O2)) - ΔGf(M203)
= (2 × 0 kJ/mol) + (1/2 × 0 kJ/mol) - (-6.00 kJ/mol)
= 12.00 kJ/mol + 0 kJ/mol + 6.00 kJ/mol
= 18.00 kJ/mol
Therefore, the standard change in Gibbs energy for the reaction as written in the forward direction is 18.00 kJ/mol.
2.) The equilibrium constant (K) of the reaction at 298K can be determined using the equation:
ΔG° = -RTln(K)
Given that ΔG° = ΔG°f, we can rearrange the equation to solve for K:
K = e^(-ΔG°/RT)
Substituting the values, where R is the ideal gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin (298 K):
K = e^(-18.00 kJ/mol / (8.314 J/(mol·K) × 298 K))
= e^(-18000 J/mol / (2.471272 J/(mol·K)))
= e^(-7280.88)
Using a scientific calculator or software to compute e^(-7280.88), we find:
K ≈ 2.95 × 10^(-3170)
Therefore, the equilibrium constant (K) of the reaction in the forward direction at 298K is approximately 2.95 × 10^(-3170).
3.) To determine the equilibrium pressure of O2(g) over M(s) at 298K, we can use the ideal gas law and the partial pressure of O2 gas. The ideal gas law is given by:
PV = nRT
Since n (number of moles) and R (ideal gas constant) are constant, we can rearrange the equation to solve for pressure (P):
P = nRT/V
However, since we don't have the volume (V) of the system, we'll focus on the equilibrium pressure of O2(g) alone. According to the stoichiometry of the reaction, for every 1/2 mole of O2 produced, 1 mole of M is consumed.
Hence, at equilibrium, the mole ratio of O2 to M is 1/2:1.
Assuming an initial pressure of O2(g) is p atm, the pressure of M(s) can be considered negligible. Therefore, at equilibrium, the partial pressure of O2(g) is given by:
P(O2,eq) = (1/2) × p
However, to calculate the equilibrium pressure of O2(g), we need to determine the value of p. For this, we can use the equilibrium constant (K) and the relationship between K and equilibrium pressures.
For the given reaction:
2M + 1/2O2 ⇌ M203
The equilibrium constant (K) is equal to the ratio of the product of the equilibrium pressures of the products over the product of the equilibrium pressures of the reactants. In this case, there is only one gaseous reactant (O2) and one gaseous product (O2), so we have:
K = P(O2,eq) / (P(M,eq))^2
Substituting the values:
2.95 × 10^(-3170) = (1/2) p / (0)^2
As the equilibrium constant is extremely small, we can take the approximation that (P(M,eq))^2 ≈ 0.
This implies that (1/2) p ≈ 0, which further implies that p ≈ 0.
Therefore, the equilibrium pressure of O2(g) over M(s) at 298K is approximately 0 atm.