The equilibrium constant for HI decomposition at 500 °C is 5.8×10-3.

2HI(g) > H2(g) + I2(g)

At that temperature reactants and products are gaseous. (There is no change in the total number of moles, so Kc = Kp.)
A sample of 0.49 mol of HI is placed in a 1.00 L vessel which is then heated to 500 °C. When equilibrium is reached, what is the molar concentration of H2? (Give units).
[H2] = 0.032 M

Now I cant figure out how to fine the concentrations of I2 and HI.
I keep getting for I2=.035M and for HI=.46M

3 answers

see above.

that question is super hard

To find the molar concentrations of I2 and HI at equilibrium, we can use the information given: the equilibrium constant (Kc) and the initial amount of HI.

Step 1: Write the balanced chemical equation
2HI(g) ⇌ H2(g) + I2(g)

Step 2: Set up an ICE table (Initial, Change, Equilibrium) to monitor the changes in concentration.

| 2HI | H2 | I2
-------------------------------------------
Initial| 0.49 | 0 | 0
Change| -2x | +x | +x
Equil. | 0.49-2x | x | x

Here, x represents the change in concentration from the initial amount of HI.

Step 3: Write the expression for the equilibrium constant (Kc).
Kc = ([H2] * [I2])/([HI]^2) = 5.8×10^-3

Step 4: Substitute the equilibrium concentrations into the Kc expression.
5.8×10^-3 = (x * x) / (0.49 - 2x)^2

Step 5: Solve the equation for x using quadratic equation or approximation method.
By solving the equation, we find that x ≈ 0.032 M.

Step 6: Calculate the equilibrium concentrations of H2 and I2.
[H2] = x = 0.032 M (as given in the question)
[I2] = x = 0.032 M

So, the molar concentrations of H2 and I2 at equilibrium are both 0.032 M.