A 1.000 L vessel is filled with 1.000 mole of N2 ,2.000 moles of H2, and 3.000 moles of NH3.When the reaction N2(g) + 3 H2(g)⇀↽2 NH3(g) comes to equilibrium, it is observed that the concentration of NH3is 2.12 moles/L. What is the numerical value of the equilibrium constant Kc?
Ah, equilibrium constants. They're like the celebrities of chemistry – always in the spotlight. Now, let's see. We know the reaction equation is N2(g) + 3 H2(g) ⇀↽ 2 NH3(g), and we're given the concentrations of NH3 at equilibrium.
To find the equilibrium constant Kc, we need to use the formula:
Kc = [NH3]² / ([N2] * [H2]³),
where [NH3], [N2], and [H2] are the concentrations of NH3, N2, and H2, respectively. Given that the concentration of NH3 is 2.12 moles/L, we'll plug that in as [NH3].
Now, we need to find the concentrations of N2 and H2. The total volume of the vessel is 1.000 L, and we have 1.000 mole of N2 and 2.000 moles of H2. Therefore, the concentrations are:
[N2] = 1.000 mol / 1.000 L = 1.000 mol/L
[H2] = 2.000 mol / 1.000 L = 2.000 mol/L
Now let's put those values into the formula:
Kc = (2.12 mol/L)² / ((1.000 mol/L) * (2.000 mol/L)³)
And voilà, after some calculations, the numerical value of the equilibrium constant Kc should be waiting for you. Good luck!
To calculate the equilibrium constant (Kc) for the given reaction:
N2(g) + 3 H2(g) ⇌ 2 NH3(g)
We can use the equilibrium concentration values of N2, H2, and NH3 along with the stoichiometric coefficients of the reaction.
Given:
Initial concentrations:
[N2] = 1.000 mole / 1.000 L = 1.000 M
[H2] = 2.000 moles / 1.000 L = 2.000 M
[NH3] = 3.000 moles / 1.000 L = 3.000 M
Concentration at equilibrium:
[NH3] = 2.12 moles / 1.000 L = 2.12 M
Using the equilibrium expression, Kc, which is defined as the ratio of the product concentrations to reactant concentrations, where each concentration is raised to the power of its stoichiometric coefficient:
Kc = [NH3]^2 / ([N2] * [H2]^3)
Substituting the given equilibrium concentration values, we get:
Kc = (2.12)^2 / (1.000 * (2.000)^3)
Simplifying:
Kc = 4.4944 / 16
Kc ≈ 0.2809
Therefore, the numerical value of the equilibrium constant Kc is approximately 0.2809.
To determine the numerical value of the equilibrium constant Kc for the given reaction, we need to first write down the balanced chemical equation and then use the given information about the initial amounts and equilibrium concentration of NH3.
The balanced chemical equation is:
N2(g) + 3 H2(g) ⇌ 2 NH3(g)
From the equation, we can see that the stoichiometric ratio between N2 and NH3 is 1:2. Therefore, if x moles of N2 react, then 2x moles of NH3 will be formed.
Let's set up an ICE table to determine the equilibrium concentrations:
| | N2(g) | H2(g) | NH3(g) |
|------|---------|---------|----------|
| Initial | 1.000 | 2.000 | 3.000 |
| Change | - x | -3x | +2x |
| Equilibrium | 1.000 - x | 2.000 - 3x | 2.12 |
Since the equilibrium concentration of NH3 is given as 2.12 moles/L, we can write:
2.12 = 2x
x = 2.12/2
x = 1.06
Now we can substitute the value of x into the expressions for the equilibrium concentrations of N2 and H2:
[N2] = 1.000 - 1.06 = -0.06 (since x is smaller than the initial amount of N2, the concentration becomes negative)
[H2] = 2.000 - 3(1.06) = -0.18
Next, we can use the equilibrium concentrations to determine the equilibrium partial pressures, which are directly proportional to the concentrations at constant temperature and volume.
To find the partial pressure of NH3, we use the ideal gas law:
PV = nRT
The pressure (P) is given by the ideal gas law as:
P = n/V
Given that the volume (V) is 1.000 L and the moles of NH3 (n) is 2.12 moles, we can calculate the pressure:
P(NH3) = 2.12/1.000 = 2.12 atm
Knowing the equilibrium partial pressures, we can use the expression for the equilibrium constant Kc:
Kc = [NH3]^2 / ([N2] * [H2]^3)
Plugging in the values:
Kc = (2.12)^2 / ((-0.06) * (-0.18)^3)
Kc ≈ 1.06 x 10^6
So, the numerical value of the equilibrium constant Kc is approximately 1.06 x 10^6.
.........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.