Some N2 and H2 are placed in an empty 5.00 liter container at 500 degree celcius. When equilibrium is established, 3.0mol N2, 2.40mol H2 and 0.565mol of NH3 are present. Evaluate Kc for the reaction at 500 degree celcius.
To evaluate Kc for the reaction, we need to write a balanced equation first. From the information given, we know that:
N2 + 3H2 ⇌ 2NH3
The balanced equation shows that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Now, let's determine the initial concentrations of the reactants and products:
N2 initial concentration = 3.0 mol / 5.00 L = 0.60 M
H2 initial concentration = 2.40 mol / 5.00 L = 0.48 M
NH3 initial concentration = 0.565 mol / 5.00 L = 0.113 M
Since each coefficient in the balanced equation represents the stoichiometric ratio, the equilibrium concentrations can be expressed in terms of x, where x is the change in concentration.
Therefore, at equilibrium:
[N2] = (0.60 - x) M
[H2] = (0.48 - 3x) M
[NH3] = (0.113 + 2x) M
Now, we can express the equilibrium constant expression, Kc, using the equilibrium concentrations:
Kc = ([NH3]^2) / ([N2] * [H2]^3)
Substituting the equilibrium concentrations into the equation:
Kc = [(0.113 + 2x)^2] / [(0.60 - x) * (0.48 - 3x)^3]
Since we have the equilibrium concentrations and the chemical equation, we can substitute the values into the equation and solve for Kc. However, we need the value of x, which can be determined by using the balanced equation and the stoichiometric ratios.
From the balanced equation, we know that 1 mole of N2 reacts with 3 moles of H2, and 2 moles of NH3 are produced. Therefore, the amount of NH3 produced is twice the amount of N2 that reacts.
Using this information, we can calculate x:
x = 2 * (0.60 M - 0.113 M) = 0.974 M
Substituting the value of x into the equation for Kc:
Kc = [(0.113 + 2 * 0.974)^2] / [(0.60 - 0.974) * (0.48 - 3 * 0.974)^3]
After substituting the values and performing the calculation, the final value of Kc can be determined.