Refer to the equilibrium system below at 500 degrees Celsius.
N20(g) + NO2(g) <-> 3NO(g)
a) only reactants are placed in a rigid container. The initial partial pressure of N2O(g) is 1.1 atm and that of NO2 is 1.9 atm. What is the partial pressure of NO(g) when the system reaches equilibrium?
b) find the standard free energy change, delta G, for the system at 500 degrees Celsius. Include units in your answer
For #1 you need Kp or Kc for the reaction OR you need the equilibrium pressure of either of the reactants at equilibriium. See below.
2. dG = -RT*ln K
#1. Here is how you do the problem. I assume you have Kp. If not let me know what you have. You can convert Kc to Kp and if you have either product's pressure at equilibiium you can calculate Kp and go from there.
.....................N20(g) + NO2(g) <-> 3NO(g)
I.......................1.1........1.9..................0
C.......................-p.........-p................3p
E....................1.1- p.....1.9 - p...........3p
Kp = (3p)^3/(1.1-p)(1.9-p)
Solve for p = pressure of NO @ equilibrium,
Post your work if you get stuck.
I forgot to include that kp= 6.6 x 10^-6 at 500 degrees C!
I assume you can take it from here. If the cubic equation gives you trouble you can find several cubic equation problem solvers by looking on Google.
To solve part (a) and find the partial pressure of NO(g) at equilibrium, we need to use the concept of equilibrium constant (Kp) and the equation for the given reaction:
N2O(g) + NO2(g) ↔ 3NO(g)
Step 1: Write the equilibrium expression using the partial pressures of the gases:
Kp = ([NO]^3) / ([N2O][NO2])
Step 2: Substitute the given initial partial pressures into the equilibrium expression:
Kp = ([NO]^3) / ([N2O][NO2])
= (x^3) / (1.1 * 1.9)
Where 'x' represents the partial pressure of NO(g) at equilibrium.
Step 3: Use the given initial partial pressures to solve for 'x' using the Kp expression.
Kp = (x^3) / (1.1 * 1.9)
Simplifying, we have:
x^3 = Kp * (1.1 * 1.9)
Step 4: Calculate 'x' by taking the cube root of both sides of the equation:
x = (Kp * (1.1 * 1.9))^(1/3)
Substitute the known values for Kp (which can be determined experimentally or from tables) and solve for 'x' to find the partial pressure of NO at equilibrium.
For part (b), to calculate the standard free energy change (ΔG) for the system at 500 degrees Celsius, we need to use the relationship between ΔG and the equilibrium constant (Kp).
ΔG = -RT ln(Kp)
Where:
ΔG is the standard free energy change,
R is the gas constant (8.314 J/(mol·K)),
T is the temperature in Kelvin (500 °C = 773 K),
ln is the natural logarithm,
and Kp is the equilibrium constant.
Substitute the values into the equation and solve for ΔG. Make sure to include the appropriate units (Joules) in your answer.