Need help in AP chemistry on Equilibrium
When heated, hydrogen sulfide gas decomposes according to the equation
2 H2S(g) ⇄ 2 H2(g) + S2(g)
A 3.40 g sample of H2S(g) is introduced into an evacuated rigid 1.25 L container. The sealed
container is heated to 483 K, and 3.72 x 10
-2
mol of S2(g) is present at equilibrium.
(a) Write the expression for the equilibrium constant, Kc, for the decomposition reaction
represented above.
(b) Calculate the equilibrium concentration, in mol L
-1
, of the following gases in the container
at 483 K.
(i) H2(g)
(ii) H2S (g)
(c) Calculate the value of the equilibrium constant, Kc, for the decomposition reaction at 483 K.
(d) For the reaction H2(g) + ½ S2(g) ↔ H2S(g) at 483 K, calculate the value of the equilibrium
constant, Kc.
(e) Calculate the partial pressure of S2(g) in the container at equilibrium at 483 K
a. mols H2S = grams/molar mass = 3.40/34 = 0.1
Surely I don't need to write the Kc expression for you.
b.
..........2H2S ==> 2H2 + S2
initial..0.1........0.....0
change....2x.......2x.....x
equil...0.1-2x.....2x.....x
From the problem, (S2) = x = 0.0372.
Therefore, (H2) is 2x
(H2S) = 0.1-2x
You can do the arithmetic.
c. Substitute the values of concns into Kc and solve for Kc.
d. The problem asks for the original Kc but half of it and reversed. Take square root to take care of the 1/2 part and take the reciprocal to take care of the reversed part.
e. Use PV = nRT.
Post your work if you get stuck.
Note that I dropped a - sign here on the change of -2x. The equil value of 0.1-2x is correct as written.
b.
..........2H2S ==> 2H2 + S2
initial..0.1........0.....0
change....2x.......2x.....x
equil...0.1-2x.....2x.....x
(a) The expression for the equilibrium constant, Kc, for the decomposition reaction is calculated using the concentrations of the products and reactants at equilibrium. In this case, the expression is:
Kc = ([H2]^2[S2]) / [H2S]^2
Where [H2], [S2], and [H2S] represent the molar concentrations of H2, S2, and H2S, respectively.
(b) To calculate the equilibrium concentration of H2(g) and H2S(g) at 483 K, we need to use the information given. We know that 3.72 x 10^-2 mol of S2(g) is present at equilibrium. Let's calculate the molar concentration for each gas using their respective coefficients:
(i) H2(g)
The coefficient of H2(g) is 2. Since the reaction is in a rigid container, the volume remains constant at 1.25 L. So the concentration of H2 is the same as the number of moles of H2 divided by the volume:
[H2] = (2 * 3.72 x 10^-2 mol) / 1.25 L
(ii) H2S(g)
The coefficient of H2S(g) is 1. Using the same logic as above, the concentration of H2S is:
[H2S] = (3.40 g / molar mass of H2S) / 1.25 L
(c) To calculate the value of the equilibrium constant, Kc, we need to substitute the molar concentrations into the equilibrium constant expression. Use the values obtained in part (b) to calculate the equilibrium constant.
(d) For the reaction H2(g) + ½ S2(g) ↔ H2S(g) at 483 K, the equilibrium constant, Kc, is different from the Kc calculated in part (c). Here, we are looking at a different reaction with different coefficients. Use the same logic to calculate the value of Kc for this reaction.
(e) To calculate the partial pressure of S2(g) in the container at equilibrium at 483 K, we need to know the total pressure in the container. If the container is evacuated, the initial pressure is 0. At equilibrium, the partial pressure of S2(g) will be determined by the equilibrium constant and the partial pressures of the other gases.
To answer these questions, you'll need to use the concepts of equilibrium, concentration, and the ideal gas law. Here's how you can approach each part:
(a) The expression for the equilibrium constant, Kc, is determined by writing the ratio of the products to the reactants, with each concentration raised to the power of its coefficient in the balanced chemical equation. In this case, the expression for Kc is:
Kc = ([H2]²[S2]) / [H2S]²
(b) To calculate the equilibrium concentration of gases in the container, you'll need to use the information given. Start by calculating the initial concentration of H2S(g) using the molar mass and the mass given in grams. Then, use the ideal gas law to convert moles to concentration by dividing by the volume of the container. For H2(g), you can use stoichiometry to relate it to the known concentration of S2(g) at equilibrium.
(c) To calculate the value of the equilibrium constant, Kc, use the equilibrium concentrations you obtained in part (b) and substitute them into the expression for Kc.
(d) For the reaction H2(g) + ½ S2(g) ↔ H2S(g) at 483 K, the equilibrium constant, Kc, is the reciprocal of the equilibrium constant for the opposite reaction (according to Le Chatelier's principle). So, if the value of Kc for the decomposition reaction is X, then the value of Kc for the reverse reaction is 1/X.
(e) To calculate the partial pressure of S2(g) at equilibrium, you can use the ideal gas law to convert moles to partial pressure. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Rearrange the equation to isolate P and then substitute the known values to find the partial pressure of S2(g).
Remember to convert all temperatures to Kelvin by adding 273.15 to the given temperature. You may also need to convert units if necessary.
I hope this guide helps you to solve the AP Chemistry equilibrium questions! If you have any specific doubts or need further assistance, feel free to ask.