# Calculate the concentration of each species at equilibrium?

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Given the initial concentration of HBr, does the same rule still applies in terms of ICE tables? I used the ICE TABLE but I am unable to calculate the concentrations.

Given the initial concentration of HBr, does the same rule still applies in terms of ICE tables? I used the ICE TABLE but I am unable to calculate the concentrations.

##### 1 Answer

Here's what I got.

#### Explanation:

I'm not sure what you mean by

... the same rule still applies in terms of ICE tables?

but **yes**, an ICE table is what you need to use here in order to find the answer.

The problem provides you with the **initial concentrations** of hydrogen gas,

Moreover, you also know the value of the **equilibrium constant**,

Notice that you have **forward reaction** will be favored, i.e. you should expect the equilibrium concentration of hydrogen bromide to be **higher** than the equilibrium concentrations of hydrogen gas and bromine gas.

Before moving on to the calculations, use the given values to calculate the **reaction quotient**, *direction* in which the equilibrium will proceed when these concentrations of reactants and product are mixed together.

#Q_c = (["HBr"]_t)/(["H"_2]_t * ["Br"_2]_t) -> # does notuseequilibrium concentrations!

In your case, you have

#Q_c = 1.20^color(red)(2)/(0.300 * 0.150) = 32#

Now, because **shift to the left**, i.e. the *reverse reaction* will be favored. As a result, the equilibrium concentrations of hydrogen gas and of bromine gas will be **higher** than their respective initial values.

Similarly, the equilibrium concentration of hydrogen bromide will be **lower** than its initial value.

Set up an **ICE table** using the initial concentrations

#" " "H"_ (2(g)) " "+" " "Br"_ (2(g)) rightleftharpoons " "color(red)(2)"HBr"_ ((g))#

By definition, the equilibrium constant will be equal to

#K_(eq) = (["HBr"]^color(red)(2))/(["H"_2] * ["Br"_2])#

In your case, this will be equal to

#K_(eq) = (1.20 - color(red)(2)x)^color(red)(2)/((0.300 + x) * (0.150 + x))#

#11.2 = (1.44 - 4.8 * x + 4x^2)/((0.045 + 0.450 * x + x^2)#

This will be equivalent to

#0.504 + 5.04 * x + 11.2 * x^2 = 1.44 - 4.8 * x + 4x^2#

Rearrange to *quadratic form* to find

#7.2 * x^2 + 9.84 * x -0.936 = 0#

Now, this quadratic equation will produce two solutions, one positive and one negative.

#x_1 = -1.456" "# and#" "x_2 = 0.0893#

Notice that in order to avoid having **negative** equilibrium concentrations for hydrogen gas and bromine gas, you must pick the *positive* value of

This means that the **equilibrium concentrations** of the three chemical species that take part in the reaction will be

#["H"_2] = 0.300 + 0.0893 = color(green)(bar(ul(|color(white)(a/a)color(black)("0.389 M")color(white)(a/a)|)))#

#["Br"_2] = 0.150 + 0.0893 = color(green)(bar(ul(|color(white)(a/a)color(black)("0.293 M")color(white)(a/a)|)))#

#["HBr"] = 1.20 - color(red)(2) * 0.0893 = color(green)(bar(ul(|color(white)(a/a)color(black)("1.02 M")color(white)(a/a)|)))#

The values are rounded to three **sig figs**.

Notice that the equilibrium concentration of hydrogen bromide is **still** higher than the equilibrium concentrations of hydrogen gas and bromine gas, as you would expect for an equilibrium for which