The equilibrium constant (Kc) for the gas phase reaction 2 NH3 --> N2 + 3 H2 is 3 x 10-3 at some temperature. The reaction is started by placing a sample of ammonia in an empty one liter flask. When equilibrium is established there is 0.010 mole of N2 present. How much ammonia was originally added to the flask?
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If the pressure on the equlibrium system 2CO(g)+O2(g)<--->2CO2(g) is increased, the quantity of CO2(g) remains the same. True or False
To solve this problem, we need to use the equilibrium expression and the given information. The equilibrium expression for the reaction is:
Kc = [N2][H2]^3 / [NH3]^2
Given:
Kc = 3 x 10^-3
[N2] = 0.010 mol/L
We are asked to find the initial amount of ammonia, [NH3].
Let's assume x mol of ammonia was originally added to the flask. At equilibrium, since two moles of NH3 react to form one mole of N2, the amount of NH3 remaining will be (x - 2 * [N2]).
Now, we can substitute the known values into the equilibrium expression:
3 x 10^-3 = (0.010)^1 * ([H2]^3) / (x - 2 * 0.010)^2
Now, we can solve for x by rearranging the equation and solving for x:
3 x 10^-3 = (0.010 * [H2]^3) / (x - 0.020)^2
(x - 0.020)^2 = (0.010 * [H2]^3) / (3 x 10^-3)
(x - 0.020)^2 = 10 * [H2]^3
Since we don't have the concentration of H2, we cannot solve for x exactly. However, we can make an approximation assuming that the reaction has reached equilibrium, the concentration of H2 is much greater compared to NH3.
Therefore, we can ignore the concentration change of H2 and assume [H2]^3 is constant. We can further simplify the equation:
(x - 0.020)^2 = Kc / 10
Taking the square root of both sides:
x - 0.020 = √(Kc / 10)
Solving for x:
x = 0.020 + √(Kc / 10)
Substituting the given value of Kc:
x = 0.020 + √(3 x 10^-3 / 10)
x = 0.020 + √(3 x 10^-4)
x = 0.020 + √(3 x 10^-4)
x = 0.020 + 0.00548
x = 0.02548 mol
Therefore, approximately 0.02548 mol of ammonia was originally added to the flask.
To find out how much ammonia was originally added to the flask, we can use the equilibrium constant (Kc) and the concentrations of the reactants and products at equilibrium.
First, let's define the initial moles of ammonia (NH3) as 'x'.
We know that for every 2 moles of ammonia, 1 mole of nitrogen gas (N2) and 3 moles of hydrogen gas (H2) are produced. So, at equilibrium, the moles of N2 would be equal to '0.010 mole' (as given in the question).
Using this information, we can determine the concentrations of NH3, N2, and H2 at equilibrium:
[NH3] = (x - 2 * 0.010) / 1 (Since 2 moles of NH3 produces 1 mole of N2)
[N2] = 0.010 / 1 (Given)
[H2] = 3 * 0.010 / 1 (Since 2 moles of NH3 produces 3 moles of H2)
Now, using the equilibrium constant (Kc), we can set up the expression:
Kc = [N2]^1 / [NH3]^2
Substituting the values, we get:
3 × 10^(-3) = (0.010 / 1)^1 / [(x - 2 * 0.010) / 1]^2
Simplifying this equation:
3 × 10^(-3) = 0.010 / (x - 0.020)^2
Now, we can solve for 'x' by rearranging the equation:
(x - 0.020)^2 = 0.010 / (3 × 10^(-3))
(x - 0.020)^2 = 100/3
Taking the square root of both sides:
x - 0.020 = √(100/3)
x = 0.020 + √(100/3)
Calculating the value of 'x':
x ≈ 0.020 + 5.77
x ≈ 5.79
Therefore, approximately 5.79 moles of ammonia were originally added to the flask.