# The equilibrium constant for the reaction is #49#. What is the equilibrium concentration of #"HI"# if #0.500# mol #"H"_2# and #0.500# mol #"I"_2# were mixed in a #"1.00-L"# container initially?

##
#"H"_2(g) + "I"_2(g) rightleftharpoons 2"HI"(g)#

Select one:

a. 0.39 M

b. 0.78 M

c. 0.22 M

d. 0.11 M

e. 0.89 M

Select one:

a. 0.39 M

b. 0.78 M

c. 0.22 M

d. 0.11 M

e. 0.89 M

You know that at an unspecified temperature, the following equilibrium reaction

has an equilibrium constant equal to

By definition, the equilibrium constant is equal to

In other words, you have

So at equilibrium, the reaction vessel will contain

and

The equilibrium constant will thus be equal to

Take the square roots of both sides to get--remember, we're looking for concentration here, so you can discard the negative solution.

This means that the equilibrium concentration of hydrogen iodide will be

The answer should be rounded to three sig figs, the number of sig figs you have for your values, but since the options given to you are rounded to two sig figs, you can say that

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The equilibrium concentration of "HI" is 7.00 mol/L.

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The balanced equation for the reaction is:

[H_2(g) + I_2(g) \rightleftharpoons 2HI(g)]

Given that the initial concentrations of (H_2) and (I_2) are both 0.500 mol in a 1.00 L container, the initial concentration of each reactant is 0.500 mol/L. Since no (HI) is initially present, its initial concentration is 0 mol/L.

Let (x) represent the change in concentration of (H_2), (I_2), and (HI).

After the reaction reaches equilibrium, the concentration of (H_2) and (I_2) will decrease by (x) mol/L, and the concentration of (HI) will increase by (2x) mol/L due to the stoichiometry of the reaction.

So, at equilibrium, the concentrations are:

[ [H_2]*{eq} = 0.500 - x \text{ mol/L}]
[ [I_2]*{eq} = 0.500 - x \text{ mol/L}]
[ [HI]_{eq} = 2x \text{ mol/L}]

The equilibrium constant expression for this reaction is given by:

[ K_c = \frac{[HI]*{eq}^2}{[H_2]*{eq} \times [I_2]_{eq}}]

Given that (K_c = 49), we can set up the equation:

[49 = \frac{(2x)^2}{(0.500 - x)^2}]

Solving this equation for (x), we find (x = 0.224).

Thus, at equilibrium:

[ [HI]_{eq} = 2x = 2 \times 0.224 = 0.448 \text{ mol/L}]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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