For the following reaction at a certain temperature:
2( ) 2( ) ( ) 2 H F HF g g g + �
it is found that the equilibrium concentrations in a 5.00 L rigid container are [H2]=0.0500 M,
[F2]=0.0100 M, and [HF]=0.400 M. If 0.200 mol F2 is added to this equilibrium mixture,
calculate the concentrations of all gases once equilibrium is reestablished.
I don't know what the gibberish you wrote stands for. If I ignore that, we can go to the problem which is
......H2 + F2 ==> 2HF--is that right?
E..0.05...0.01....0.4
Ka = (H2)(F2)/(HF^2)
Plug in the E concns and solve for Ka. Then 0.2 mol F added which is 0.2 mol/5 L = 0.04M so initial for F2 = 0.01+0.04 0.05M
I.....0.05..0.05...0.40
C.......-x...-x.....+2x
E.....0.05-x.0.05-x..0.40+2x
Substitute E line into Ka and solve x then evaluate HF, H2 and F2.
C
To solve this problem, we can use the stoichiometry of the reaction and the concept of equilibrium to determine the concentrations of the gases at equilibrium.
First, let's write down the balanced equation for the reaction:
2 H2 (g) + F2 (g) ⇌ 2 HF (g)
According to the given information, the initial concentrations of H2, F2, and HF are:
[H2] = 0.0500 M
[F2] = 0.0100 M
[HF] = 0.400 M
Now, let's calculate the change in concentration of F2 when 0.200 mol is added. Since the stoichiometry of the reaction is 1:1 for H2 and F2, the change in concentration of F2 will also be 0.200 M.
Next, let's determine the change in concentration of HF. Since the stoichiometry of the reaction is 2:1 for HF and F2, the change in concentration of HF will be twice the change in concentration of F2. Therefore, the change in concentration of HF is 2 * 0.200 M = 0.400 M.
Now, let's calculate the new equilibrium concentrations. The concentration of H2 will remain unchanged since it is not directly involved in the change. Therefore, [H2] = 0.0500 M.
The new concentration of F2 can be calculated by adding the change in concentration to the initial concentration:
[F2] = 0.0100 M + 0.200 M = 0.2100 M.
The new concentration of HF can be calculated by subtracting the change in concentration from the initial concentration:
[HF] = 0.400 M - 0.400 M = 0.000 M.
Therefore, the concentrations of all gases at equilibrium are:
[H2] = 0.0500 M
[F2] = 0.2100 M
[HF] = 0.000 M.
To answer this question, we need to use the principles of equilibrium and the concept of Le Chatelier's principle.
1. Write the balanced chemical equation for the reaction:
2 H2(g) + F2(g) ⇌ 2 HF(g)
2. Write the expression for the equilibrium constant, Kc:
Kc = [HF]^2 / ([H2]^2 * [F2])
3. Calculate the initial concentration of HF, [HF]0, using the given equilibrium concentration:
[HF]0 = 0.400 M
4. Calculate the initial concentration of H2, [H2]0, using the given equilibrium concentration:
[H2]0 = 0.0500 M
5. Calculate the initial concentration of F2, [F2]0, using the given equilibrium concentration:
[F2]0 = 0.0100 M + 0.200 mol / 5.00 L = 0.0500 M
Note: We added 0.200 mol of F2 to a 5.00 L container, so the increase in concentration is (0.200 mol / 5.00 L) = 0.0400 M.
6. Substitute the initial concentrations into the Kc expression:
Kc = [HF]0^2 / ([H2]0^2 * [F2]0)
7. Solve for Kc using the given values:
Kc = (0.400 M)^2 / ((0.0500 M)^2 * 0.0500 M)
8. Calculate the change in concentration for H2, [H2], using Le Chatelier's principle:
[H2] = [H2]0 - 2x
9. Calculate the change in concentration for F2, [F2], using Le Chatelier's principle:
[F2] = [F2]0 - x
10. Calculate the change in concentration for HF, [HF], using Le Chatelier's principle:
[HF] = [HF]0 + 2x
11. Substitute the equilibrium concentrations into the Kc expression:
Kc = [HF]^2 / ([H2]^2 * [F2])
12. Substitute the given values and solve the quadratic equation to find the value of x:
Kc = (0.400 M + 2x)^2 / ((0.0500 M - 2x)^2 * (0.0500 M - x))
Solving for x will provide the change in concentration for H2 and F2.
13. Calculate the equilibrium concentrations of H2, F2, and HF using the values of x obtained in step 12:
[H2] = [H2]0 - 2x
[F2] = [F2]0 - x
[HF] = [HF]0 + 2x
14. Substitute the values into the equations and calculate:
[H2] = 0.0500 M - 2x
[F2] = 0.0100 M + 0.0400 M - x = 0.0500 M - x
[HF] = 0.400 M + 2x
15. Use the value of x obtained in step 12 to calculate the equilibrium concentrations:
[H2] = 0.0500 M - 2x
[F2] = 0.0500 M - x
[HF] = 0.400 M + 2x