# Question 60f23

May 19, 2015

The equilibrium constant for this reaction will be equal to 3.13.

So, you know that you're dealing with a generic equilibrium that involves three species, $A$, $B$, both reactants, and $C$, the product.

Even before doing any calculations, you can predict that ${K}_{c}$ will be bigger than 1. Notice that the concentration of $A$ decreases, while the concentration of $C$ increases.

This tells you that the initial concentrations will cause the equilibrium to shift right, i.e. the reaction quotient, ${Q}_{c}$, is smaller than ${K}_{c}$.

If this is the case, the equilibrium concentration of $B$ will decrease as well, but not by the same amount as the concentration of $A$ - this happens because of the difference between the stoichiometric coefficients of the two reactants.

So, you were only given two equilibrium concentrations, more specifically those of $A$ and of $C$. However, you can use the initial concentrations to determine what the equilibrium concentration of $B$ would be.

To do that, use an ICE table for you equilibrium reaction

$\text{ " " "A " "+ " "color(red)(2)B " " rightleftharpoons " } C$
I......0.650............1.20...............0.700
C.......(-x)..............(-$\textcolor{red}{2}$x)...................(+x)
E.....0.650-x........1.20-2x..........700+x

You know that ${\left[A\right]}_{\text{equilibrium" = "0.450 M}}$. This means that

$0.650 - x = 0.540 \implies x = 0.200$

Check to see if it matches the equilibrium concentration of $C$

${\left[C\right]}_{\text{equilibrium" = 0700 + x = 0.700 + 0.200 = "0.900 M}}$

This means that the equilibrium concentration of $B$ will be

${\left[B\right]}_{\text{equilibrium" = 1.2 - 2x = 1.2 - 2 * 0.200 = "0.800 M}}$

Therefore, the equilibrium constant for this reaction will be

K_c = ([C])/([A] * [B]^color(red)(2)) = 0.900/(0.450 * 0.800"^2) = color(green)(3.13)#

SIDE NOTE The value of ${Q}_{c}$ at the start of the reaction was 0.74, which again confirms our initial prediction.