# 0.2 mole of A and 0.4 mile of B were reacted at a certain temperature and allowed to come to equilibrium . A+B---2AB The equilibrium mixture contained 0.1 mole of A, find Kc if volume of the container was 2 dm^3?

Apr 28, 2017

$\textcolor{red}{{K}_{c} = 13.4}$

#### Explanation:

I'll show my explanation in steps for your understanding

1. First set balance the equation

$2 A + 2 B = 2 A B$

2. Now write the equilibrium constant for the reaction

$\textcolor{b l u e}{{K}_{c} = \frac{\left[A B\right]}{\left[A\right] \left[B\right]}}$

3. As $\textcolor{b l u e}{{K}_{c}}$ is being determined the temperature matters and thus volume also matters.So the conversion of moles to molarity is necessary

$M = \frac{\text{moles}}{L}$

4. As the volume of the container is ${\text{2dm}}^{3}$

Conversionof ${\mathrm{dm}}^{3}$ to litres is necessary

$1 {\mathrm{dm}}^{3} = 1 L$

$\therefore$ $2 {\mathrm{dm}}^{3} = 2 L$

5. Calculate the molarities

Conc. of A = $\frac{\text{0.2moles}}{2 L} = 0.1 M$

Conc. of B = $\frac{\text{0.4mole}}{2 L} = 0.2 M$

Conc. of A at equilibrium $\frac{\text{0.1moles}}{2 L} = 0.05 M$

5. Set up an ICE table

$\textcolor{b l u e}{\text{For problems related to equilibrium}}$
$\textcolor{b l u e}{\text{concentrations,you need to set up an ICE table.}}$

color(white)(mmmmml)"2A" color(white)(lm)+ color(white)(mmmm)"2B"color(white)(llll) → color(white)(ml) 2AB
$\text{I/mol:} \textcolor{w h i t e}{m l l} 0.1 M \textcolor{w h i t e}{m m m m m l} 0.2 M \textcolor{w h i t e}{m m m m m l} 0$
$\text{C/mol:} \textcolor{w h i t e}{m} - x \textcolor{w h i t e}{m m m m m m} - x \textcolor{w h i t e}{m m m m} + 2 x$
$\text{E/mol:} \textcolor{w h i t e}{l m} 0.1 M - x \textcolor{w h i t e}{m m l} 0.2 M - x \textcolor{w h i t e}{l m m m l} 2 x$

From the given information we know that the equilibrium amount of A is 0.05M. Thus $0.1 M - x = 0.05 M$

Solve for $x$

$0.1 M - x = 0.05 M$
$x = 0.1 M - 0.05 M$
$x = 0.05 M$

6. Solve for the equilibrium amounts of B and AB

$A B = 2 x = 2 \times 0.05 M = 0.1 M$
$B = 0.2 M - 0.05 M = 0.15 M$
And we know the equilibrium amounts of A

7. Now solve for ${K}_{c}$

${K}_{c} = \text{products"/"reactants}$

${K}_{c} = \frac{\left[0.1 M\right]}{\left[0.05 M\right] \left[0.15 M\right]}$

${K}_{c} = 13.333$

If you want $\textcolor{b l u e}{\text{three sig figs}}$

${K}_{c} = 13.4$

You can also calculate the ${Q}_{c}$ to see to which direction will the reaction proceed

$\textcolor{b l u e}{{Q}_{c} = \frac{{\left[A B\right]}^{2}}{{\left[A\right]}^{2} {\left[B\right]}^{2}}}$

${Q}_{c} = 177.777777778$

As $\textcolor{p u r p \le}{{K}_{c} < {Q}_{c}}$

Reaction will proceed towards left