At 500 ºC, the decomposition of water into hydrogen and oxygen,
2H2O(g) 2H2(g) + O2(g)
has Kc = 6.0 × 10-28. What are the concentrations of H2 and O2 that are present at equilibrium in a 2.00 L reaction vessel at this temperature if the container originally held 0.086 mol H2O?
(H2O( = 0.0860 mol/2.00L = approx 0.043M
............2H2O --> 2H2 + O2
Substitute the E line into the Kc expression and solve for x.
To solve this problem, we need to use the given equilibrium constant (Kc) and the initial amount of water (H2O) to find the concentrations of H2 and O2 at equilibrium.
Step 1: Write the balanced equation:
2H2O(g) -> 2H2(g) + O2(g)
Step 2: Use the stoichiometry of the equation to determine the change in the number of moles:
Δn(H2) = 2 x (-x) = -2x
Δn(O2) = (-x)
Step 3: Define the equilibrium concentrations in terms of x:
[H2] = 0.086 - 2x
[O2] = x
Step 4: Substitute the equilibrium concentrations into the equation for Kc:
Kc = [H2]^2 * [O2] / [H2O]^2
6.0 × 10^(-28) = (0.086 - 2x)^2 * x / (0.086)^2
Step 5: Solve the equation for x. Rearrange the equation to isolate x:
(0.086 - 2x)^2 * x / (0.086)^2 = 6.0 × 10^(-28)
(0.086 - 2x)^2 * x = 6.0 × 10^(-28) * (0.086)^2
Step 6: Solve for x using algebra or a numerical method. In this case, we will use a numerical method or a graphing calculator to find x ≈ 2.8 × 10^-19.
Step 7: Use the value of x to find the concentrations of H2 and O2 at equilibrium:
[H2] = 0.086 - 2x ≈ 0.086
[O2] = x ≈ 2.8 × 10^-19
Therefore, at equilibrium, the concentrations of H2 and O2 in the 2.00 L reaction vessel are approximately:
[H2] = 0.086 mol/L
[O2] = 2.8 × 10^-19 mol/L
To find the equilibrium concentrations of H2 and O2 at 500 ºC, we can use the given equilibrium constant (Kc) and the initial moles of H2O.
First, let's calculate the initial amount of moles of H2 and O2. From the balanced equation, we see that for every 2 moles of H2O, we get 2 moles of H2 and 1 mole of O2.
Initial moles of H2O = 0.086 mol
Number of moles of H2 = 2 * 0.086 mol = 0.172 mol
Number of moles of O2 = 0.086 mol
Now, let's convert the moles of H2 and O2 to their respective concentrations (in mol/L) using the given reaction vessel volume.
Volume of reaction vessel = 2.00 L
Concentration of H2 (at equilibrium) = Number of moles of H2 / Volume of reaction vessel
Concentration of H2 = 0.172 mol / 2.00 L = 0.086 M
Concentration of O2 (at equilibrium) = Number of moles of O2 / Volume of reaction vessel
Concentration of O2 = 0.086 mol / 2.00 L = 0.043 M
Therefore, at equilibrium, the concentrations of H2 and O2 in the 2.00 L reaction vessel at 500 ºC are 0.086 M and 0.043 M, respectively.