The equilibrium constant Kc for the reaction N2(g) + 3H2(g) -> <- 2NH3 at 450 degrees celcius is 0.159. Calculate the equilibrium compostion when 1.00 mol N2 is mixed with 3.00 mol H2 in a 5.00-L vessel.
.............N2 + 3H2 ==> 2NH3
begin(mols)..1.0..3.00.....0
change........-x...-3x.....+2x
final.......1.0-x..3.00-3x..+2x
Kc = (NH3)^3/(N2)(H2)^3
Substitute the ICE chart values into the Kc expression and solve for x and the other values. (x is i moles).
Then moles/5.00 L = M
Note the correct spelling of celsius.
To calculate the equilibrium composition when 1.00 mol N2 is mixed with 3.00 mol H2 in a 5.00-L vessel, we need to use the given equilibrium constant (Kc) and the stoichiometry of the balanced equation.
The balanced equation for the reaction is:
N2(g) + 3H2(g) -> 2NH3(g)
First, we need to calculate the initial concentrations of N2 and H2. To do this, we divide the number of moles of each gas by the volume of the vessel:
For N2: [N2] = moles of N2 / volume of vessel = 1.00 mol / 5.00 L = 0.20 M
For H2: [H2] = moles of H2 / volume of vessel = 3.00 mol / 5.00 L = 0.60 M
Next, we'll use the equilibrium constant (Kc) to calculate the equilibrium concentrations.
Kc is defined as the ratio of the product concentrations to the reactant concentrations, each raised to their respective stoichiometric coefficients.
Kc = ([NH3]^2) / ([N2] * [H2]^3)
Since we have the value of Kc, we can rearrange the equation to solve for the concentration of NH3:
[NH3]^2 = Kc * ([N2] * [H2]^3)
[NH3]^2 = 0.159 * (0.20 * (0.60)^3)
[NH3]^2 = 0.159 * (0.20 * 0.216)
[NH3]^2 = 0.00650208
[NH3] = √0.00650208
[NH3] ≈ 0.0806 M
Thus, the equilibrium concentration of NH3 is approximately 0.0806 M.
To summarize the equilibrium composition:
[N2] = 0.20 M
[H2] = 0.60 M
[NH3] = 0.0806 M