8. The following partial pressure were measured at equilibrium at 2270C. pNH3 = 3atm , pN2 =
6atm and pH2 =4atm. Calculate KP and KC at 2270C for the reaction : 2NH3(g) N2 (g)
+ 3H2(g
To calculate KP and KC at 2270C for the given reaction, you need to use the equilibrium partial pressures of the reactants.
First, let's write the balanced equation for the reaction:
2NH3(g) ⇌ N2(g) + 3H2(g)
To calculate KP, which is the equilibrium constant in terms of partial pressures, you can use the formula:
KP = (pN2 × pH2^3) / pNH3^2
Substituting the given values into the formula:
KP = (6 atm × (4 atm)^3) / (3 atm)^2
KP = (6 × 64) / 9
KP = 384 / 9
KP ≈ 42.67 (rounded to two decimal places)
To calculate KC, which is the equilibrium constant in terms of molar concentrations, you can use the formula:
KC = (cN2 × cH2^3) / cNH3^2
However, we do not have the concentrations of the reactants, only their partial pressures. To calculate KC, we need to convert the partial pressures to concentrations using the ideal gas law.
The ideal gas law states: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.
Rearranging the ideal gas law equation, we get:
n = (PV) / RT
Since we have partial pressures, we can assume constant volume (V) and temperature (T), and we can use the equation:
n = P / RT
Now, we need to find the moles of each gas at equilibrium:
nNH3 = pNH3 / RT = 3 atm / (0.0821 L∙atm/(mol∙K) × 227 + 273) ≈ 0.121 mol
nN2 = pN2 / RT = 6 atm / (0.0821 L∙atm/(mol∙K) × 227 + 273) ≈ 0.243 mol
nH2 = pH2 / RT = 4 atm / (0.0821 L∙atm/(mol∙K) × 227 + 273) ≈ 0.162 mol
Now that we have the moles of each gas, we can calculate KC using the formula:
KC = (cN2 × cH2^3) / cNH3^2
Substituting the values:
KC = ((nN2 / V) × (nH2 / V)^3) / (nNH3 / V)^2
KC = (nN2 × nH2^3) / nNH3^2
KC = (0.243 × (0.162)^3) / (0.121)^2
KC ≈ 0.348 (rounded to three decimal places)
So, at 2270C, the equilibrium constants are approximately KP = 42.67 and KC = 0.348.