can anyone explain how to do this problem?
At 673 K, the equilibrium constant for the
reaction
2 HI(g) ⇀↽ H2(g) + I2(g)
is Kc = 0.0200. If 0.5 mol of HI is placed in
a 2.00 liter container and heated to 673 K,
what concentration of H2 will be present at
equilibrium?
Answer in units of M.
To find the concentration of H2 at equilibrium, we need to use the given equilibrium constant (Kc) and the initial concentration of HI. Here are the steps to solve the problem:
Step 1: Write the balanced equation:
2 HI(g) ⇀↽ H2(g) + I2(g)
Step 2: Define the initial concentrations:
Given that 0.5 mol of HI is placed in a 2.00 liter container, we can calculate the initial concentration of HI:
Initial concentration of HI = (moles of HI) / (volume of the container) = 0.5 mol / 2.00 L = 0.25 M
Step 3: Set up the equilibrium expression:
The equilibrium expression for this reaction is:
Kc = [H2]^2/[HI]^2
Step 4: Substitute the known values into the equilibrium expression:
Kc = (x)^2 / (0.25)^2
Here, 'x' is the concentration of H2 at equilibrium.
Step 5: Solve for 'x':
Rearranging the equation, we have:
Kc * (0.25)^2 = x^2
x^2 = Kc * (0.25)^2
x = √(Kc * (0.25)^2)
Step 6: Calculate the concentration of H2 at equilibrium:
Substitute the values into the equation:
x = √(0.0200 * (0.25)^2)
x ≈ 0.0071 M (rounded to four significant figures)
Therefore, the concentration of H2 at equilibrium is approximately 0.0071 M.
(HI) = 0.5mol/2L = 0.25 M
................. 2HI(g) ⇀↽ H2(g) + I2(g)
I..................0.25M.........0...........0
C,,,,,,,,,,,,,,,,,,-2x.............x............x
E,,,,,,,,,,,,,9,25-x..............x............x
Kc = (H2)(I2)/(HI)^2
Plug in the E line and solve for x = (H2) = (I2)
Post your work if you get stuck.