Question

This reaction has an equilibrium constant of \(K_{\mathrm{p}}=2.2 \times 10^{6}\) at 298 \(\mathrm{K}\) \begin{equation}2 \mathrm{COF}_{2}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{CF}_{4}(g)\end{equation} Calculate \(K_{\mathrm{p}}\) for each reaction and predict whether reactants or products will be favored at equilibrium. \begin{equation}\begin{array}{l}{\text { a. } \operatorname{COF}_{2}(g) \rightleftharpoons \frac{1}{2} \mathrm{CO}_{2}(g)+\frac{1}{2} \mathrm{CF}_{4}(g)} \\ {\text { b. } 6 \mathrm{COF}_{2}(g) \rightleftharpoons 3 \mathrm{CO}_{2}(g)+3 \mathrm{CF}_{4}(g)} \\ {\text { c. } 2 \mathrm{CO}_{2}(g)+2 \mathrm{CF}_{4}(g) \rightleftharpoons 4 \mathrm{COF}_{2}(g)}\end{array}\end{equation}

Short answer

\(K_{p_{a}}\approx 1.5 \times 10^{3}, K_{p_{b}} \approx 1.1 \times 10^{19}, K_{p_{c}} \approx 2.1 \times 10^{-13}\). For all reactions, products will be favored due to the original Kp being greater than 1.

Step-by-step solution

Understand Equilibrium Constants

The equilibrium constant, expressed as Kp, indicates the extent of a reaction; if Kp is much larger than 1, the products are favored. Conversely, if Kp is much smaller than 1, the reactants are favored at equilibrium. We're given the Kp for the first reaction and need to find the Kp for the other reactions listed.

Relating Reaction Quotients to the Changes in Reaction

When a reaction is multiplied by a coefficient, the Kp for the new reaction becomes the original Kp raised to the power of that coefficient. When a reaction is reversed, the new Kp becomes the reciprocal of the original Kp.

Calculate Kp for Reaction a

For reaction a, which is half the original reaction, take the square root of the original Kp: \(K_{p_{a}} = (K_{p_{original}})^{1/2} = (2.2 \times 10^{6})^{1/2}\).

Calculate Kp for Reaction b

For reaction b, which is three times the original reaction, raise the original Kp to the power of 3: \(K_{p_{b}} = (K_{p_{original}})^{3} = (2.2 \times 10^{6})^{3}\).

Calculate Kp for Reaction c

For reaction c, which is the reverse of the original reaction multiplied by 2, take the reciprocal of the original Kp and raise it to the power of 2: \(K_{p_{c}} = (1/K_{p_{original}})^{2} = (1/(2.2 \times 10^{6}))^{2}\).

Predict the Direction of Equilibrium

Compare the calculated Kp values with 1 to determine whether reactants or products are favored. For Kp much larger than 1, products are favored; for Kp smaller than 1, reactants are favored.

Key concepts

Chemical Equilibrium

When we talk about chemical equilibrium in the context of a chemical reaction, imagine a busy city intersection where cars are coming and going, but the number of cars on each street stays constant over time. Similarly, in a chemical reaction at equilibrium, the rate at which the reactants turn into products is exactly balanced by the rate at which the products revert back to reactants.

This doesn't mean that the reactions have stopped; rather, they're occurring simultaneously and at equal rates, so the concentrations of reactants and products remain unchanged. It's essential to understand that equilibrium does not imply that the amounts of reactants and products are equal, but that their ratios do not change over time.

Let's take a simple example: A + B ⇌ C + D. In this case, at equilibrium, the rate at which A and B combine to form C and D is the same as the rate at which C and D break down to form A and B. The equilibrium constant (Kp in the case of gases, or Kc for concentrations) gives us the ratio of these concentrations at equilibrium, providing insight into the relative amounts of each species present in the system.

Kp Calculation

Understanding Kp in Gas Reactions

When dealing with reactions involving gases, the equilibrium constant is expressed in terms of partial pressures, and we represent it as Kp. Calculating Kp is straight-forward when you know the balanced chemical equation of the reaction at equilibrium and the partial pressures of the gases involved.

In general, the Kp expression is derived from the balanced equation and looks something like this: for the reaction aA(g) + bB(g) ⇌ cC(g) + dD(g), the Kp will be expressed as \(K_{p} = \frac{{P^{c}_{C} \times P^{d}_{D}}}{{P^{a}_{A} \times P^{b}_{B}}}\) where P stands for the partial pressure of each gas.

To calculate Kp for a new reaction scenario, as mentioned in the exercise, we may need to raise Kp to a power, take its reciprocal, or take its root, depending on the changes to the coefficients or the direction of the reaction.

Reaction Quotient

The reaction quotient, Q, is a concept closely related to the equilibrium constant K. It serves as a 'snapshot' of a reaction at any given moment, rather than at equilibrium. To calculate Q, we use the same expression as K, but with the current partial pressures or concentrations of the reactants and products.

If Q = K, the system is at equilibrium. When Q < K, there are too many reactants; the reaction will proceed forward to produce more products. Conversely, if Q > K, there are too many products, and the reaction will shift to form more reactants.

By comparing Q to K, chemists can predict which way a reaction will need to shift to reach equilibrium. This comparison can be particularly useful during the initial stages of a reaction or when changing conditions, like pressure or temperature, that might drive a system away from its established equilibrium.