Question 89c43

Apr 30, 2017

C. Pressure will decrease because $Q < {K}_{\textrm{c}}$.

Explanation:

To decide the direction in which an equilibrium reaction will go, you must calculate $Q$ and compare it with ${K}_{\textrm{c}}$.

The two expressions are similar, but ${K}_{\textrm{c}}$ uses equilibrium concentrations, while $Q$ uses initial concentrations.

We have

$\textcolor{w h i t e}{m m m m m l} {\text{PCl"_3(g) + "Cl"_2(g) ⇌ "PCl}}_{5} \left(g\right)$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m} 1.5 \textcolor{w h i t e}{m m m l l} 1.0 \textcolor{w h i t e}{m m m l l} 2.5$

${K}_{\textrm{c}} = \left(\left[{\text{PCl"_5])/(["PCl"_3]["Cl}}_{2}\right]\right) = 6.5$

Q = (["PCl"_5])/(["PCl"_3]["Cl"_2]) = 2.5/(1.5 ×1.0) = 1.7#

$Q < {K}_{\textrm{c}}$. What does this mean?

We usually talk of ${K}_{\textrm{c}}$ as "products over reactants".

The system has "just enough products" to be at equilibrium.

If $Q < {K}_{\textrm{c}}$, there aren't "enough products" to reach equilibrium.

The system must react to create more products, that is, it must move to the right.

However, every time we create one mole of product, we remove two moles of reactants.

The number of particles in the flask decreases, so the pressure in the flask must also decrease.