# Question 964ef

Nov 1, 2017

Here's what I got.

#### Explanation:

• ["A"] = "0.650 M"
• ["B"] = "1.15 M"
• ["C"] = "0.500 M"

and that, at equilibrium, the reaction vessel contains

• ["A"] = "0.530 M"
• ["C"] = "0.620 M"

Now, notice that the concentration of $\text{A}$ decreased since you have

$\text{0.530 M " - " 0.650 M" = -"0.120 M}$

The minus sign tells you that the concentration of $\text{A}$ decreased by $\text{0.120 M}$.

By comparison, the concentration of $\text{C}$ increased because you have

$\text{0.520 M " - " 0.620 M " - " 0.500 M" = "0.120 M}$

This is consistent with the fact that the reaction proceeded in the forward direction, i.e. the forward reaction was favored. This means that you can represent the reaction as

${\text{A"_ ((aq)) + color(red)(2)"B"_ ((aq)) -> "C}}_{\left(a q\right)}$

You can also say that because the forward reaction is favored, the equilibrium constant must come out to be $> 1$.

As you can see, for every $1$ mole of $\text{A}$ that takes part in the reaction, the reaction consumes $\textcolor{red}{2}$ moles of $\text{B}$ and produces $1$ mole of $\text{C}$.

So if the reaction consumed $\text{0.120 M}$ of $\text{A}$ and produced $\text{0.120 M}$ of $\text{C}$, it follows that it must have consumed

$\textcolor{red}{2} \cdot \text{0.120 M" = "0.240 M}$

of reactant $\text{B}$. You can thus say that the concentration of $\text{B}$ decreased by $\text{0.240 M}$, which implies that, at equilibrium, the reaction vessel will contain

["B"] = "1.15 M " - " 0.240 M" = "0.910 M"

By definition, the equilibrium constant for this reaction will take the form

${K}_{c} = \left({\left[\text{C"])/(["A"] * ["B}\right]}^{\textcolor{red}{2}}\right)$

Now all you have to do is plug in the equilibrium concentrations of the chemical species that take part in the reaction. Since all three chemical species are in the aqueous state, they will be included in the expression of ${K}_{c}$.

For the sake of simplicity, I'll leave the answer without added units!

${K}_{c} = \frac{0.620}{0.530 \cdot {\left(0.910\right)}^{\textcolor{red}{2}}} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{1.41}}}$

I'll leave the answer rounded to three sig figs, but keep in mind that you should have

["B"] = "1.15 M " - " 0.120 M" = "0.91 M"#

The equilibrium concentration of $\text{B}$ should be rounded to two decimal places, which means that you have two sig figs for $0.91$.

This would make

${K}_{c} = \frac{0.620}{0.530 \cdot {\left(0.91\right)}^{\textcolor{red}{2}}} = 1.4$

As predicted, the value of the equilibrium constant is indeed $> 1$.