A 1.000 L vessel is filled with 2.000 moles of

N2, 1.000 mole of H2, and 2.000 moles of NH3.
When the reaction
N2(g) + 3 H2(g) ⇀↽ 2 NH3(g)
comes to equilibrium, it is observed that the
concentration of H2 is 2.53 moles/L. What is
the numerical value of the equilibrium constant
Kc?

Well, in this reaction, we have nitrogen holding hands with hydrogen to form ammonia. Quite a friendly gathering, don't you think?

Now, to find the equilibrium constant, we need to use the law of mass action, which states that the ratio of the concentrations of the products to the concentrations of the reactants is constant at equilibrium. Mathematically, we can write it as:

Kc = [NH3]^2 / ([N2] * [H2]^3)

Here, the square brackets represent the molar concentrations of the respective species.

Given that the equilibrium concentration of H2 is 2.53 moles/L, we can substitute the values into the equation:

Kc = [(2.000)^2] / [(2.53) * (2.000)^3]

Now, let me calculate that for you...

*scribbling on a virtual chalkboard*

Okay, after crunching the numbers, the equilibrium constant, Kc, is approximately 0.313.

So, let's raise our beakers and toast to the friendly gathering of nitrogen, hydrogen, and ammonia! Cheers!

To find the numerical value of the equilibrium constant, Kc, for the given reaction N2(g) + 3 H2(g) ⇀↽ 2 NH3(g), we can use the equilibrium concentrations of the reactants and products.

Given:
Initial volume of the vessel = 1.000 L
Initial moles of N2 = 2.000 moles
Initial moles of H2 = 1.000 mole
Initial moles of NH3 = 2.000 moles
Concentration of H2 at equilibrium = 2.53 moles/L

From the balanced equation, we know that the coefficient of H2 is 3. This means that for every 3 moles of H2, 2 moles of NH3 are formed.

To calculate the equilibrium concentrations, we need to find the change in moles of H2 and NH3 (Δn), which can be calculated as follows:

Δn(H2) = Initial moles of H2 - moles of H2 at equilibrium
= 1.000 mole - 2.53 moles/L * 1.000 L
= 1.000 mole - 2.53 moles
= -1.53 moles

Δn(NH3) = 2 * Δn(H2)
= 2 * (-1.53) moles
= -3.06 moles

Now, we can calculate the equilibrium concentrations of the reactants and products:

Concentration of N2 at equilibrium = (Initial moles of N2 + Δn(N2)) / Initial volume of the vessel
= (2.000 moles) / (1.000 L)
= 2.000 moles/L

Concentration of H2 at equilibrium = (Initial moles of H2 + Δn(H2)) / Initial volume of the vessel
= (1.000 mole + (-1.53 moles)) / (1.000 L)
= (-0.53) moles/L

Concentration of NH3 at equilibrium = (Initial moles of NH3 + Δn(NH3)) / Initial volume of the vessel
= (2.000 moles + (-3.06 moles)) / (1.000 L)
= (-1.06) moles/L

Now, we can write the expression for the equilibrium constant, Kc, using the concentrations of the reactants and products:

Kc = ([NH3]^2) / ([N2] * [H2]^3)

Substituting the equilibrium concentrations, we get:

Kc = ((-1.06) moles/L)^2 / (2.000 moles/L * (-0.53) moles/L^3)

Calculating the numerical value of Kc:

Kc = 1.12 L^2/mole^2

Therefore, the numerical value of the equilibrium constant (Kc) is 1.12 L^2/mole^2.

To find the numerical value of the equilibrium constant Kc, we need to determine the concentrations of all the species in the reaction at equilibrium.

Given:
Initial volume (V) = 1.000 L
Initial moles of N2 (n1) = 2.000 mol
Initial moles of H2 (n2) = 1.000 mol
Initial moles of NH3 (n3) = 2.000 mol
Concentration of H2 at equilibrium (C2) = 2.53 mol/L

From the balanced chemical equation:
N2(g) + 3 H2(g) ⇀↽ 2 NH3(g)

The stoichiometry of the reaction tells us that for every 3 moles of H2 reacted, 2 moles of NH3 are formed. Therefore, the change in the number of moles of H2 (Δn2) would be (3 * -x), and the change in the number of moles of NH3 (Δn3) would be (2 * x), where 'x' is the amount reacted or formed at equilibrium.

At equilibrium, the moles of H2 (n2') and NH3 (n3') can be calculated using the initial moles and the change in moles, as follows:

n2' = n2 - Δn2 = n2 - (3 * -x) = n2 + 3x
n3' = n3 + Δn3 = n3 + (2 * x) = n3 + 2x

Since the volume (V) remains constant at 1.000 L, the concentration (C) can be calculated using the formula:

C = n/V

We can express the equilibrium concentrations of H2 (C2') and NH3 (C3') as:

C2' = n2' / V
C3' = n3' / V

Given that C2' = 2.53 mol/L, we can substitute the values to the equations above:

2.53 mol/L = (n2 + 3x) / 1.000 L
C3' = (n3 + 2x) / 1.000 L

Now, we can solve the equations to find the value of 'x':

From the equation for C2':
2.53 = (1.000 + 3x) / 1.000

Simplifying:
2.53 = 1 + 3x

Rearranging the equation:
3x = 2.53 - 1
3x = 1.53

Dividing by 3:
x = 0.51

Now, we have the value of 'x', which represents the number of moles of H2 reacted or formed at equilibrium. Since there are 3 moles of H2 in the balanced equation, we can calculate the moles of H2 remaining as (1 - x):

n2' = 1.000 mol - 0.51 mol = 0.49 mol

Finally, to find the equilibrium constant Kc, we can use the formula:

Kc = (C3')^2 / (C2')^3

Substituting the values:
Kc = (0.49 mol / 1.000 L)^2 / (2.53 mol / 1.000 L)^3

Calculating the value of Kc:
Kc = 0.2427 / (2.53^3)
Kc = 0.2427 / 16.18
Kc ≈ 0.015

Therefore, the numerical value of the equilibrium constant Kc is approximately 0.015.

.........N2(g) + 3 H2(g) ⇀↽> 2 NH3(g)

I.........2M........1M..........2M
C........-2x........-3x.........2x
E..........?.......2.53M.........?

Since we know eq (H2) = 2.53, that means that H2 was formed, the reaction is proceeding from right to left and that means we must change the sign of the C line. I will do that here.
........N2(g) + 3 H2(g) ⇀↽> 2 NH3(g)
I.......2M........1M..........2M
C......+2x........+3x.........-2x
E.......?.......2.53M..........?

Then we know 1 + 3x = 2.53 which means x must be 0.51
When you know x you can calculate all of the C line. So equilibrium N2 must be 0.51 x 2 and add 2 and equilibrium NH3 must be 0.51 x 2 and that subtracted from 2M. Plug into Keq expression and solve for Kc.