Question

At a certain temperature, \(K_{\mathrm{c}}=0.18\) for the equilibrium \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) Suppose a reaction vessel at this temperature contained these three gases at the following concentrations: \(\left[\mathrm{PCl}_{3}\right]=\) \(0.0420 \mathrm{M},\left[\mathrm{Cl}_{2}\right]=0.0240 \mathrm{M},\left[\mathrm{PCl}_{5}\right]=0.00500 \mathrm{M}\) (a) Compute the reaction quotient and use it to determine whether the system is in a state of equilibrium. (b) If the system is not at equilibrium, in which direction will the reaction proceed to get to equilibrium?

Short answer

Q is calculated as 4.34, which is greater than Kc (0.18), indicating that the reaction will proceed to the left towards reactants to reach equilibrium.

Step-by-step solution

Determine Reaction Quotient (Q)

Calculate the reaction quotient (Q) by substituting the given concentrations into the expression for Q, which is based on the balanced equation. For the given reaction, the expression for Q is: Q = [PCl5] / ([PCl3][Cl2]). Substitute the concentrations to get Q.

Calculate Reaction Quotient (Q)

Substitute the given concentration values into the Q expression: Q = ([PCl5]) / ([PCl3][Cl2]) = (0.00500 M) / (0.0420 M * 0.0240 M).

Compare Q to Kc to Determine Equilibrium State

Compare the calculated Q value to the Kc value. If Q = Kc, the system is at equilibrium. If Q < Kc, the reaction will proceed to the right (towards products) to reach equilibrium. If Q > Kc, the reaction will proceed to the left (towards reactants) to reach equilibrium.

Conclude Direction of Reaction

Once the comparison is made between Q and Kc, conclude whether the system is at equilibrium or the direction in which the reaction needs to proceed to achieve equilibrium.

Key concepts

Understanding Reaction Quotient (Q)

Imagine you're in a race, and the reaction quotient (Q) is the measure of how far you've gone compared to the total distance – the finish line being chemical equilibrium. For the reaction of phosphorus trichloride and chlorine gas forming phosphorus pentachloride, we look at the concentrations of the gases involved to calculate Q. In this race, we check if we're ahead, behind, or right at the finish line.

In our reaction vessel, we know the concentrations of each gas: \begin{align*}\[\text{PCl}_{3}\] &= 0.0420 \text{ M},\[\text{Cl}_{2}\] &= 0.0240 \text{ M},\[\text{PCl}_{5}\] &= 0.00500 \text{ M}.\end{align*} To find Q, we plug these numbers into our race tracker formula: \begin{equation*}Q = \frac{\[\text{PCl}_{5}\]}{\[\text{PCl}_{3}\]\times \[\text{Cl}_{2}\]}.\end{equation*} Let’s do the math: \begin{equation*}Q = \frac{0.00500}{0.0420 \times 0.0240} \approx 4.96.\end{equation*}

What Does Q Reveal?

- When Q equals the equilibrium constant (Kc), our race is over; we've reached equilibrium.
- A Q < Kc indicates we are 'behind', and the reaction must move forward (towards products) to reach equilibrium.
- A Q > Kc means we are 'ahead', so the reaction moves in reverse (towards reactants) to find balance.
In our case, Q significantly exceeds our given Kc of 0.18. Still, it's like we've sprinted past the equilibrium finish line. According to this, our reaction needs to backtrack to reach that sweet spot – equilibrium.

Equilibrium Constant (Kc) Explained

The equilibrium constant (Kc) is the race referee; it reports the perfect finish line position at a specific temperature. For our equilibrium reaction between PCl3, Cl2, and PCl5, we've been told that Kc is 0.18 at the reaction temperature. Compare this constant to a rulebook value; it doesn't change unless the conditions, like temperature, are altered.

The value of Kc is determined through experimentation and tells us about the recipe for equilibrium. A small Kc (much less than 1) means the finish line favours the reactants – in our analogy, it's a short race. A large Kc (much greater than 1) suggests the products are the stars, and our race's finish line is far down the track, indicating a product-heavy equilibrium mix.

Connecting Kc and Q:

- If Q < Kc, there's too much reactant, and not enough product. The reaction 'runs' to make more product.
- If Q > Kc, there's an excess of product, and natural tendencies correct this by forming more reactant.
- At Q = Kc, nobody moves; it's perfect equilibrium. Relax, the race is done.
Knowing that, when we have our calculated Q significantly higher than the given Kc of 0.18, we interpret that the reaction is not at equilibrium. It's like our runners (the reaction) need to head back to decrease the product concentration.

Navigating Shifts with Le Châtelier's Principle

Le Châtelier's principle is the coach that predicts how our reaction runners will respond to strategy changes. It states: When a chemical system at equilibrium is disturbed, it will shift in a direction that counteracts the change. This principle supports the idea of balance, hinting that the equilibrium mix doesn't enjoy being tipped one way or the other.

An alteration in concentration, temperature, or pressure can prompt a shift to re-establish equilibrium. For example, adding more product is like telling the runners they've gone too far; the reaction adjusts by moving towards the reactants.

Real-Life Application:

If the concentration of one of the reactants is increased in our reaction, the principle predicts a shift towards more products to even things out. Conversely, if we take away some product, the reaction will ramp up product formation to balance the loss.
When dealing with pressure changes for gas reactions, remember this: increasing pressure favours the side with fewer gas molecules, while decreasing pressure favours the side with more.
Le Châtelier’s principle guides us when predicting the outcome of disruptions, letting us maintain not only chemical harmony but also clarity for students tackling the dynamic world of chemical reactions.