At 462degrees, the reaction
(1) heat + 2 NOCl(g) <=====> 2 NO(g) + Cl2(g)
has an equilibrium constant, Keq = 8.0 x 10-2.
a) What is Keq at 462degrees for the reaction
2 NO(g) + Cl2(g) <=====> 2 NOCl(g)
1/Keq
may I ask how did you get this answer?
Keq for the reverse reaction is simply the reciprocal.
ok but What if the equation was then NOCl <=> NO + 1/2Cl2 ?
Then Keq for the reaction as written is sq rt original Keq.
It works this way.
2A + B2 ==> 2C Keq = 10
A + 1/2 B2 ==> C (Keq)1/2 = sq rt 10
4A + 2B2 ==> 4C (Keq)2 = 100
2C ==> 2A + B2 Keq = 0.1
Well, in that case, Keq for the reaction NOCl <=> NO + 1/2Cl2 would be the square root of the original Keq. It's kind of like taking the square root of 10. So, Keq for the reaction as written would be the square root of the original Keq.
In the case of the equation NOCl <=> NO + 1/2Cl2, the Keq for the reaction as written is the square root of the original Keq. This is because the stoichiometric coefficients in the balanced equation are different.
To explain this further, let's take a look at the relationship between Keq and the balanced equation for a reaction:
Original equation: 2A + B2 <=> 2C
Keq = 10
If we divide the equation by 2, we get:
A + 1/2 B2 <=> C
The new equation has different stoichiometric coefficients than the original equation. In this case, the Keq for the new equation is the square root of the original Keq:
(Keq)^(1/2) = sqrt(Keq) = sqrt(10)
Similarly, if we multiply the original equation by 2, we get:
4A + 2B2 <=> 4C
(Keq)^2 = 100
And if we reverse the original equation, we get:
2C <=> 2A + B2
Keq = 0.1
In summary, the Keq for a reaction can be adjusted by changing the stoichiometric coefficients in the balanced equation, and in some cases, by taking the square root or reciprocal of the original Keq value.
In this case, if the equation is NOCl <=> NO + 1/2Cl2, the Keq for the reaction as written is the square root of the original Keq.
To understand this, let's consider the balanced equation and its corresponding equilibrium expression for the given reaction:
NOCl <=> NO + 1/2Cl2
The equilibrium expression for this reaction would be:
Keq = [NO][1/2Cl2] / [NOCl]
Now, if we compare this equation with the original equation:
2 NO(g) + Cl2(g) <=> 2 NOCl(g)
We can observe that the species concentrations in the new equation are the square root of the original concentrations:
[NO] in the new equation = [NO]^(1/2)
[Cl2] in the new equation = [Cl2]^(1/2)
[NOCl] in the new equation = [NOCl]
Applying these new concentrations to the equilibrium expression, we get:
Keq_new = [NO]^(1/2) * [Cl2]^(1/2) / [NOCl]
Simplifying this expression, we get:
Keq_new = ([NO] * [Cl2])^(1/2) / [NOCl]
Since the original Keq is equal to [NO] * [Cl2] / [NOCl], we can substitute it in the equation above:
Keq_new = ([NO] * [Cl2])^(1/2) / [NOCl] = sqrt([NO] * [Cl2] / [NOCl]) = sqrt(Keq)
Therefore, the Keq for the reaction NOCl <=> NO + 1/2Cl2 is the square root of the original Keq.