Question

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}(g)+\mathrm{Cl}(g) $$ At \(25^{\circ} \mathrm{C}\), the rate constants for the forward and reverse reactions are \(1.4 \times 10^{-28} M^{-1} \mathrm{~s}^{-1}\) and \(9.3 \times 10^{10} \mathrm{M}^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C}\) ? (b) Are reactants or products more plentiful at equilibrium?

Short answer

The equilibrium constant at \(25^{\circ} \mathrm{C}\) is \(K = 1.51 \times 10^{-39}\). Since the value of K is very small, the equilibrium lies far to the left, and reactants are more plentiful at equilibrium than products.

Step-by-step solution

Write down the given information

We are given the rate constants for the forward and reverse reactions: Forward reaction rate constant, \(k_f = 1.4 \times 10^{-28} M^{-1}s^{-1}\) Reverse reaction rate constant, \(k_r = 9.3 \times 10^{10} M^{-1}s^{-1}\)

Relate rate constants to the equilibrium constant

In a chemical equilibrium, the equilibrium constant, K, can be related to the rate constants of the forward and reverse reactions using the following formula: \(K = \frac{k_f}{k_r}\) Substitute the given values of \(k_f\) and \(k_r\) into the equation.

Calculate the equilibrium constant

Substituting the given values in the equation: \(K = \frac{1.4 \times 10^{-28} M^{-1}s^{-1}}{9.3 \times 10^{10} M^{-1}s^{-1}}\) Now solve for K: \(K = 1.51 \times 10^{-39}\) Therefore, the value of the equilibrium constant at 25°C is \(1.51 \times 10^{-39}\).

Determine whether reactants or products are more plentiful at equilibrium

The magnitude of the equilibrium constant, K, gives us insight into the relative concentrations of products and reactants at equilibrium. For reactions with very small K values (K << 1), the equilibrium lies far to the left, which means that there will be mostly reactants at equilibrium. For reactions with very large K values (K >> 1), the equilibrium lies far to the right, which means that there will be mostly products at equilibrium. In our case, the value of K is very small (\(1.51 \times 10^{-39}\)), so the equilibrium lies far to the left. Therefore, reactants will be more plentiful at equilibrium than products.

Key concepts

Chemical Equilibrium

Imagine a busy street where pedestrians are walking in both directions. Now imagine a point where just as many people are heading one way as the other—that's a bit like chemical equilibrium. It's the state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, so the overall concentrations of reactants and products remain constant over time. It doesn't mean the reactions have stopped; rather, they're happening at the same rate, leading to a dynamic balance.

When we write an equation like \( CO(g) + Cl_{2}(g) \rightleftharpoons COCl(g) + Cl(g) \), this arrow suggests a system that can go both ways and may reach equilibrium. In our case, with a calculated equilibrium constant \( K = 1.51 \times 10^{-39} \), we observe a strong preference for the reactants, hinting that the system is scarcely proceeding towards products at equilibrium.

Rate Constants

Have you ever timed how fast someone runs a race? Well, in chemistry, we time reactions with something called rate constants. These constants, represented by \( k \), tell us how quickly reactants turn into products, or vice versa in a reverse reaction. They're a bit like a recipe that gives us a clue to how fast a cake bakes. If the rate constant is high, the reaction goes quickly; if it's low, the reaction takes its sweet time.

In our textbook problem, we had two rate constants: \( k_f = 1.4 \times 10^{-28} M^{-1}s^{-1} \) for the forward reaction and \( k_r = 9.3 \times 10^{10} M^{-1}s^{-1} \) for the reverse reaction. It's clear that our reverse reaction races ahead much faster than the forward reaction, and this greatly influences where the equilibrium will settle.

Elementary Steps

A dance routine is made up of individual moves, or steps—and so is a chemical reaction. Elementary steps are like these single dance moves, the simplest processes in a reaction pathway that can't be broken into simpler ones. Each step has its own rate constant, and they come together to show us the overall dance of the reaction.

In simple terms, if we consider our equilibrium equation \( CO(g) + Cl_{2}(g) \rightleftharpoons COCl(g) + Cl(g) \), each arrow represents an elementary step with its own rate constant, taking us either towards the product or back to the reactants. Understanding these steps allows chemists to piece together the broader choreography of the chemical reaction.

Reaction Rates

Let's cook with science for a bit—reaction rates measure how fast ingredients (reactants) turn into a delicious meal (products). In a chemical context, it's the speed at which a reaction proceeds. It involves looking at how the concentration of reactants or products changes over time. It's crucial to note that reaction rate is influenced by several factors, such as temperature, pressure, and the presence of a catalyst.

In the context of our original problem, the reaction rate for the forward and reverse processes determines how quickly we reach equilibrium, and as we've seen, since the reverse reaction has a much higher rate constant than the forward, it works at a much quicker pace—meaning we're likely to see less product formed when equilibrium is finally achieved.

Le Chatelier's Principle

Imagine a seesaw that's perfectly balanced. Now, put weight on one side—naturally, it tips over, right? Chemistry has its own seesaw in Le Chatelier's principle, which tells us that if a change is made to a system at equilibrium, the system will adjust itself to counteract that change and a new equilibrium is established. This could be changing the concentration, pressure, or temperature of the system.

In our exercise, if we suddenly increase the temperature or add more of a reactant, the equilibrium will shift to help absorb these extra 'hassles'—imagine the seesaw trying to balance itself again. Knowing this, chemists can tweak conditions to get the best yield of products or to understand how systems will react when disturbed.