5.88 moles of nitrogen and 16.2 moles of oxygen are mixed until the equilibrium is established, 11.28 moles of nitric oxide are formed calculate the value of equilibrium constant
N2+ O2 <=>2NO
5,88 16,2 0
0,24 10,56 11,28
K = (11,28^2)/(0,24.10,56) = 50,2
You don't have information listed. Concentration is what goes into the Keq calculation. You have no volume.
Using mols will not work. The Keq involves concentrations; mols are not concentrations.
Explanation:
Start by writing the balanced chemical equation for this equilibrium reaction
N
2(g]
+
O
2(g]
⇌
2
NO
(g]
Now, the equilibrium constant for a chemical reaction is calculated by using the equilibrium concentrations of the chemical species that take part in that reaction.
In your case, the problem provides you with moles, not with molarities. However, if you take into account the fact that all three gases share the same reaction vessel, you can use moles instead of molarities.
More on that later.
So, you know that you're mixing two gases, nitrogen gas,
N
2
, and oxygen gas,
O
2
at
2000
∘
C
, until equilibrium is established with nitric oxide,
NO
.
You start by adding
5.88
moles of
N
2
and
16.2
moles of
O
2
to the reaction vessel. Initially, the reaction vessel does not contain any nitric oxide.
However, after equilibrium is established, you find that the vessel contains
11.2
moles of nitric oxide.
This means that you can use an ICE table to find the number of moles of each reactant present at equilibrium
N
2(g]
+
O
2(g]
⇌
2
NO
(g]
I
5.88
16.2
0
C
(
−
x
)
(
−
x
)
(
+
2
x
)
E
5.88
−
x
16.2
−
x
11.2
Notice that the number of moles of nitric oxide increased by
2
x
, where
x
represents the number of moles of each reactant consumed to make the product.
This of course means that you have
2
x
=
11.2
⇒
x
=
11.2
2
=
5.6
At equilibrium, the reaction vessel will thus contain, along with
11.2
moles of
NO
5.88
−
5.6
=
0.28 moles N
2
16.2
−
5.6
=
10.6 moles O
2
Now, before doing any calculations, try to predict whether or not the equilibrium constant,
K
c
, is smaller than
1
or bigger than
1
.
As you know,
K
c
<
1
implies that the reaction favors the reactants, and
K
c
>
1
means that the raection favors the products.
In this case, the number of moles of each reactant decreased significantly. At the same time, the number of moles of the product increased significantly. This tells you that you can expect to see
K
c
>
1
.
By definition,
K
c
will be equal to
K
c
=
[
NO
]
2
[
N
2
]
⋅
[
O
2
]
Now, here is why I said that you can use the number of moles and not worry about molarity. Let's assume that the reaction vessel has a volume
V
liters. The equilibrium concentrations of the the chemical species will be
[
N
2
]
=
0.28 moles
V
L
=
0.28
V
M
[
O
2
]
=
10.6 moles
V
L
=
10.6
V
M
[
NO
]
=
11.2 moles
V
L
=
11.2
V
M
Plug these values into the expression for
K
c
to get
K
c
=
11.2
2
V
2
M
2
10.6
V
M
⋅
0.28
V
M
=
11.2
2
10.6
⋅
0.28
Therefore,
K
c
=
3.31
The answer is rounded to three sig figs.
Indeed, our prediction that
Kc> 1
To calculate the equilibrium constant (K) for this reaction, we need to use the balanced chemical equation and the stoichiometry of the reaction. The balanced equation for the reaction between nitrogen (N2) and oxygen (O2) to form nitric oxide (NO) is:
2N2 + O2 ⇌ 2NO
According to the equation, 2 moles of nitrogen react with 1 mole of oxygen to produce 2 moles of nitric oxide.
From the given information, we know that 5.88 moles of nitrogen and 16.2 moles of oxygen are mixed. Therefore, nitrogen is the limiting reactant because there is only enough nitrogen to react with 2.94 moles of oxygen.
Based on the stoichiometry, we can calculate the amount of nitric oxide formed using the limiting reactant:
- Since 2 moles of nitrogen react to produce 2 moles of nitric oxide, 5.88 moles of nitrogen will produce (5.88/2) x 2 = 5.88 moles of nitric oxide.
Therefore, the amount of nitric oxide formed is 5.88 moles.
Now we have all the necessary information to calculate the equilibrium constant (K):
K = ([products] / [reactants])^coefficients
K = ([NO]^2) / ([N2]^2 x [O2])
Plugging in the values:
K = (11.28^2) / (5.88^2 x 2.94)
K = 126.998 / 102.958
K ≈ 1.232
Therefore, the value of the equilibrium constant for this reaction is approximately 1.232.