Consider the decomposition of a metal oxide to its elements, where M represents a generic metal:
M2O3 (s) <=> 2M (s) + 3/2 O2 (g)
Given a delta Gf: -770 Kj for the M2O3
0 Kj for the M
0 Kj for the O2
What is the standard change in Gibbs energy for the reaction, as written, in the forward direction?
* I think -770 Kj ??
What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K?
* K is products over reactants which leaves me with 0 ???
What is the equilibrium pressure of O2(g) over M(s) at 298 K?
* I honestly have no idea and am not even sure how to approach this!
Any information is greatly appreciated, thank you!
Right digits; wrong sign.
dGrxn = (n*dG products) - (dG reactants) = (0) - (-770) = +770.
dG = -RT*lnK
Substitute and solve for Kp.
Kp = pO23/2
Thank you VERY much!
I still don't understand what to do after setting the K= pO2^(3/2). What do you substitute into the O2?
To calculate the standard change in Gibbs energy (ΔG) for a reaction, you can use the equation:
ΔG = ΣnΔGf(products) - ΣmΔGf(reactants)
where n and m are the stoichiometric coefficients of the products and reactants, and ΔGf is the standard Gibbs energy of formation for each component.
In this case, the reaction is:
M2O3 (s) <=> 2M (s) + 3/2 O2 (g)
Given the values:
ΔGf(M2O3) = -770 kJ
ΔGf(M) = 0 kJ
ΔGf(O2) = 0 kJ
Plugging in these values, we have:
ΔG = (2 * 0 kJ) + (3/2 * 0 kJ) - (-770 kJ) = 770 kJ
Therefore, the standard change in Gibbs energy for the reaction, as written in the forward direction, is +770 kJ.
To calculate the equilibrium constant (K) of the reaction at 298 K, you can use the equation:
ΔG = -RT ln(K)
where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and ln is the natural logarithm.
We need to convert the temperature from Kelvin to Celsius:
298 K - 273.15 = 24.85°C
Now, let's plug in the values and solve for K:
770 kJ = -8.314 J/(mol·K) * (298 K) * ln(K)
ln(K) = -0.0919
K = e^(-0.0919)
K ≈ 0.912
Therefore, the equilibrium constant of the reaction, as written in the forward direction at 298 K, is approximately 0.912.
To determine the equilibrium pressure of O2(g) over M(s) at 298 K, we need to use the expression:
K = (P(O2))^m / (P(M))^n
where P(O2) and P(M) are the partial pressures of O2(g) and M(s), respectively, and m and n are the stoichiometric coefficients in the equation.
In this case, the stoichiometric coefficients are:
m = 3/2 (for O2)
n = 2 (for M)
Since the reaction is in equilibrium, the ratio of the partial pressures is equal to the equilibrium constant:
K = P(O2)^3/2 / (P(M))^2
Let's approximate K to 0.912 and solve for P(O2) / P(M):
0.912 = (P(O2))^3/2 / (P(M))^2
Taking the square root on both sides:
0.912^(2/3) = P(O2) / P(M)
P(O2) / P(M) ≈ 0.950
Therefore, at equilibrium and at 298 K, the equilibrium pressure of O2(g) over M(s) is approximately 0.950.