A 2.00L flask was filled with 4.00 mol of HI at a certain temperature and given sufficient time to react. At equilibrium the concentration of H2 was 0.400 M. Find the equilibrium concentrations of I2 and HI and then find the Keq at this temperature 2HI(g) <-> H2(g) + I2(g)
M HI = mols/L = 4.00/2.00 = 2.00
.......2HI ==> H2 + I2
I ......2.......0....0
C.......-2x.....x....x
E.....2-2x......x....x
Substitute the E line into Keq expression. You know x = (H2) = 0.4,calculate concns and Keq from that.
To find the equilibrium concentrations of I2 and HI, we need to use stoichiometry and the given information about the concentration of H2.
First, let's define the initial concentration of HI, which is 4.00 mol in a 2.00L flask. Using the concentration formula c = n/V, we find that the initial concentration of HI is 4.00 mol/2.00 L = 2.00 M.
Since the balanced equation is 2HI(g) <-> H2(g) + I2(g), we can see that the molar ratio between HI and I2 is 2:1. This means that for every 2 moles of HI that react, 1 mole of I2 is produced.
Since the concentration of H2 is given as 0.400 M, which is also the concentration of the equilibrium, we know that for every 1 mole of H2 produced, 1 mole of I2 is produced. Therefore, the concentration of I2 at equilibrium is also 0.400 M.
To find the concentration of HI at equilibrium, we need to use the stoichiometry. Since 2 moles of HI react to produce 1 mole of H2, the decrease in concentration of HI is equal to the increase in concentration of H2. Therefore, the concentration of HI at equilibrium is 2.00 M - 0.400 M = 1.60 M.
Now that we have the equilibrium concentrations of H2, I2, and HI, we can calculate the equilibrium constant Keq using the formula Keq = [H2][I2]/[HI]^2.
Substituting the values we found: Keq = (0.400 M)(0.400 M)/(1.60 M)^2 = 0.400*0.400/2.56 = 0.0625.
Therefore, the equilibrium constant Keq at this temperature is 0.0625.