# PCl3(g) + Cl2(g) <---> PCl5(g) Kc = 0.18 In an initial mixture of 0.300M PCl3 (g), 0.400M Cl2 (g), what are the equilibrium concentrations of all gases?

Dec 20, 2014

The equilibrium concentrations are $\left[P C {l}_{3}\right] = 0.280 M$, $\left[C {l}_{2}\right] = 0.380 M$, $\left[P C {l}_{5}\right] = 0.02 M$.

One can approach this problem by using the ICE method (more here: http://en.wikipedia.org/wiki/RICE_chart); even before doing any calculations, the value of the equilibrium constant can give us an idea on how the result will look like; since ${K}_{c} < 1$, the reaction will favor the reactans, so we'd expect bigger equilibrium concentrations for $P C {l}_{3}$ and $C {l}_{2}$, than for $P C {l}_{5}$.

..$P C {l}_{3} \left(g\right) + C {l}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s P C {l}_{5} \left(g\right)$
I:. 0.300.............0.400..............0
C:... -x....................-x...................+x
E:..(0.300-x).....(0.400-x)...........x

So, ${K}_{c} = \frac{\left[P C {l}_{5}\right]}{\left[P C {l}_{3}\right] \cdot \left[C {l}_{2}\right]} = \frac{x}{\left(0.300 - x\right) \cdot \left(0.400 - x\right)} = 0.18$

The equation $0.18 {x}^{2} - 1.126 x + 0.0216 = 0$ produces two values for $x$, ${x}_{1}$ = $6.23$ and ${x}_{2}$ = $0.020$; since concentrations cannot be negative, the correct value will be $x = 0.020$ ($x = 6.23$ would have produced negative concentrations for $P C {l}_{3}$ and $C {l}_{2}$).

Therefore, the equilibrium concentrations for all the gases are

$\left[P C {l}_{3}\right] = 0.300 - 0.020 = 0.28 M$
$\left[C {l}_{2}\right] = 0.400 - 0.020 = 0.38 M$
$\left[P C {l}_{5}\right] = 0.020 M$

Notice that the initial predictions on the relative concentration values is correct, the concentrations of the reactans are bigger than the concentration of the product.