Equilibrium Constant of A(g) 2B(g)+C(g) at Temperature t

In summary: L we get concentration of 0.0125mol/L which does not make sense. In summary, the conversation discusses the equilibrium constant for a chemical reaction involving gases A, B, and C at temperature t. Using the initial concentration of A and the concentration of C at equilibrium, the concentrations of B and C can be determined. The equilibrium constant, Kc, can then be calculated using the concentrations of all three gases and their stoichiometric coefficients. The equation used is Kc = [C(g)] * [B(g)]^2 / [A(g)].
  • #1
ChemRookie
39
0
A(g) <---
<--- 2B(g) + C(g)

When 1.00mol of A is placed in a 4.00L container at temperature t, the concentration of C at equilibrium is 0.050mol/L. What is the equilibrium constant for the reaction at temperature t?

Thanks for the help.
 
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  • #2
A(g)-> 2B(g)+ C(g)
1mol
(1-x)/4 2x/4 x/4
they have given that x=0.05mol/L
kc= (0.05/4*0.01/16)/0.95/4
=1/19*100mol/L
 
  • #3
ChemRookie said:
A(g) <---
<--- 2B(g) + C(g)

When 1.00mol of A is placed in a 4.00L container at temperature t, the concentration of C at equilibrium is 0.050mol/L. What is the equilibrium constant for the reaction at temperature t?

Thanks for the help.

OK let's look at the problem. First what is the initial concentration of A. Second given the stoichiometry of the problem and the value of C at equilibrium what are the concentrations of A and B ? Finally how due you write the equilibrium constant (and don't forget powers). Due the math ;)
 
  • #4
campa said:
A(g)-> 2B(g)+ C(g)
1mol
(1-x)/4 2x/4 x/4
they have given that x=0.05mol/L
kc= (0.05/4*0.01/16)/0.95/4 =1/19*100mol/L

can u explain to me how u set that up? and how do you solve it? (do the calculations)
 
  • #5
[tex]Ka= [2x]^{2}[x]/[initial conc.-x][/tex]

Can you explain to us why it'll be in such a form?
 
  • #6
the equilibrium constant for the reaction at temperature t should be Kc so considering
the moles used to start this reaction which is 1mol of A(g) you can determine the moles used to get the right side of the equation at equilibrium. So the moles used from A(g) should be (1-x)moles and they have given that at equilibrium the concentration of C is 0.05mol(x=0.05) and it equals to the concentration of B(g) but there are two moles of B so it should be 2*0.05
and by deviding these moles by the volume of the container you get mol/L of each gas
kc= [C(g)] *[B(g)]*[B(g)]/[A(g)]

I hope this helps and my guess is that it should be 0.05mol and not 0.05mol/L
 
Last edited:

Related to Equilibrium Constant of A(g) 2B(g)+C(g) at Temperature t

1. What is the equation for calculating the equilibrium constant for this reaction?

The equilibrium constant (Kc) for this reaction is equal to the concentration of products (BeqCeq) divided by the concentration of reactant (Aeq). This can also be written as Kc = [BeqCeq] / [Aeq].

2. How does the value of the equilibrium constant change with temperature?

The value of the equilibrium constant is dependent on temperature. As the temperature increases, the value of Kc also increases. This is because an increase in temperature causes the reaction to favor the endothermic direction, leading to more products being formed.

3. How does the equilibrium constant relate to the position of equilibrium?

The equilibrium constant is a measure of the position of equilibrium. A larger value of Kc indicates that the equilibrium favors the formation of products, while a smaller value indicates that the equilibrium favors the formation of reactants.

4. What does a large equilibrium constant indicate about the reaction?

A large equilibrium constant (Kc) indicates that the reaction strongly favors the formation of products at equilibrium. This means that the reaction is proceeding in the forward direction to a greater extent.

5. How can the equilibrium constant be used to predict the direction of a reaction?

If the equilibrium constant (Kc) is larger than 1, then the reaction is considered to be product-favored and will proceed in the forward direction. If Kc is smaller than 1, then the reaction is reactant-favored and will proceed in the reverse direction. If Kc is equal to 1, then the reaction is at equilibrium and there will be no net change in the concentrations of products and reactants.

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