Kc = 28 at some temperature for the reaction
H2(g) + I2(g) ⇀↽ 2 HI(g).
If 16.7 moles of HI are introduced into a
10.0 liter vessel, how many moles of I2 are
present at equilibrium?
Answer in units of mol.
(HI) = 16.7/10 = 1.67 M
.....H2 + I2 ==> 2HI
I.....0....0......1.67
C.....x....x......-2x
E.....x....x.....1.67-2x
Plug the E line into the Kc expression and solve for x
To find the number of moles of I2 at equilibrium, we can use the balanced equation and the given value of the equilibrium constant (Kc).
The balanced equation for the reaction is:
H2(g) + I2(g) ⇀↽ 2 HI(g)
According to the stoichiometry of the reaction, the mole ratio between HI and I2 is 2:1. This means that for every 2 moles of HI, there is 1 mole of I2.
Given that 16.7 moles of HI are introduced into the vessel, we can use the mole ratio to calculate the number of moles of I2.
Number of moles of I2 = (16.7 moles HI) / (2 moles of HI per 1 mole of I2)
= 8.35 moles I2
Therefore, there are 8.35 moles of I2 present at equilibrium.
To determine the number of moles of I2 present at equilibrium, we first need to set up an expression for the equilibrium constant (Kc) of the reaction.
The equation for the reaction is:
H2(g) + I2(g) ⇀↽ 2 HI(g)
The equilibrium constant expression for this reaction is:
Kc = [HI]^2 / [H2] × [I2]
We are given that Kc = 28.
Next, we need to determine the initial concentrations of HI, H2, and I2. We are given that 16.7 moles of HI are introduced into a 10.0 liter vessel. This means the initial concentration of HI is:
[HI]initial = (moles of HI) / (volume of vessel)
[HI]initial = 16.7 mol / 10.0 L
To solve for the initial concentration of HI, we divide the moles of HI by the volume of the vessel.
Now, since the reaction is in equilibrium, the concentrations of all the substances will be the same as their initial concentrations. Therefore, the concentration of HI at equilibrium is still [HI]initial = 16.7 mol / 10.0 L.
Let's assume the number of moles of I2 at equilibrium is x mol. Since the stoichiometric coefficient for I2 in the balanced equation is 1, the concentration of I2 at equilibrium is given by [I2] = x mol / 10.0 L.
Using the equilibrium constant expression, we can now write:
Kc = [HI]^2 / [H2] × [I2]
Substituting the given values, we have:
28 = (16.7 mol / 10.0 L)^2 / [H2] × (x mol / 10.0 L)
Simplifying, we get:
28 = (2.7889 mol^2 / L^2) / [H2] × (x mol / 10.0 L)
Cross-multiplying, we have:
28 × [H2] × 10.0 L = 2.7889 mol^2 / L^2 × x mol
280 × [H2] = 2.7889 mol^2 / L^2 × x mol
Dividing both sides by (2.7889 mol^2 / L^2), we get:
x mol = (280 × [H2]) / (2.7889 mol^2 / L^2)
Now, substitute the initial concentration of HI ([HI]initial = 16.7 mol / 10.0 L) into the equation:
x mol = (280 × [H2]) / (2.7889 mol^2 / L^2)
x mol = (280 × (16.7 mol / 10.0 L)) / (2.7889 mol^2 / L^2)
Simplifying further, we get:
x mol = 474.193 mol^2 / L / 2.7889 mol^2 / L^2
x mol = 170.042 mol
Therefore, at equilibrium, the number of moles of I2 is 170.042 mol.