At a particular temperature, Kc = 54 for the reaction H2(g) + I2 (g) <=> 2 HI(g).
One mole of HI is placed in a 3.0-L container. What would be the equilibrium concentration of HI?
See below.
for this one, when we set up the ice chart, and we have the M of HI, and we have to subtract x, would we subtract 2x because the mols of HI is two?
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To find the equilibrium concentration of HI, we need to use the equilibrium constant expression, Kc, and the stoichiometry of the reaction.
First, let's write down the balanced equation and the expression for the equilibrium constant:
H2(g) + I2(g) ⇌ 2 HI(g)
The equilibrium constant expression, Kc, is equal to the concentration of products raised to their stoichiometric coefficients divided by the concentration of reactants raised to their stoichiometric coefficients.
Kc = [HI]^2 / ([H2] * [I2])
In this case, we have one mole of HI in a 3.0-L container, which means the concentration of HI is 1 mole / 3.0 L = 1/3 M (since the concentration is expressed in moles per liter).
Now, we can substitute the given values into the equilibrium constant expression to find the equilibrium concentration of HI:
Kc = (1/3)^2 / ([H2] * [I2])
We don't have the initial concentrations of H2 and I2, so let's assume their initial concentrations are x M.
Now, let's set up an ICE (Initial, Change, Equilibrium) table to help us solve for the equilibrium concentration of HI:
H2 + I2 ⇌ 2 HI
Initial: x x 1/3
Change: -x -x +2x
Equilibrium: x-x x-x 1/3+2x
Since the equilibrium concentration of HI is given as 1/3 M, we can set up the equation:
1/3 + 2x = 1/3
Simplifying the equation, we get:
2x = 0
x = 0
Therefore, the equilibrium concentration of HI is:
1/3 M.
So, when one mole of HI is placed in a 3.0-L container, the equilibrium concentration of HI would also be 1/3 M.