# Question #96bac

Mar 23, 2015

The equilibrium constant for this reaction is ${K}_{c} = 110.$

I'll assume that the initial number of moles of ${I}_{2}$ is equal to the initial number of moles of $B {r}_{2}$.

So, you know that you're dealing with an equilibrium between ${I}_{2}$ and $B {r}_{2}$ on one side, and $I B r$ on the other side. Notice that the equilibrium concentration of $I B r$ is larger than the initial concentrations of the reactans.

If you start with only reactants in the flask, the fact that you have more product at equilibrium than you had reactants inititially tells you that ${K}_{c}$ is greater than 1.

The initial concentrations of the reactants are

${C}_{{I}_{2}} = {C}_{B {r}_{2}} = \frac{n}{V} = \text{0.500 moles"/"1 L" = "0.5 M}$

To determine the actual value of the equilibrium constant, use the ICE chart method (more here: https://en.wikipedia.org/wiki/RICE_chart).

.......${I}_{2} + B {r}_{2} r i g h t \le f t h a r p \infty n s \textcolor{red}{2} I B r$
I......0.5.......0.5..............0
C...(-x).........(-x)............$\textcolor{red}{+ 2 x}$
E..0.5-x......0.5-x..........0.840

Since $2 x = 0.840$, the value of $x$ will be

$x = \frac{0.840}{2} = \text{0.420}$

This means that the equilibrium concentrations of the reactants will be

$\left[{I}_{2}\right] = \left[B {r}_{2}\right] = 0.5 - 0.420 = \text{0.08 M}$

The value of ${K}_{c}$ will be

${K}_{c} = {0.840}^{2} / \left(0.08 \cdot 0.08\right) = 110.25$

Since you've been a little inconsistent with the number of sig figs given, I'll round the answer to three sig figs (although that 1L volume would call for rounding to one sig fig)

${K}_{c} = \textcolor{red}{110.}$