Consider the reversible reaction shown below: \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g)\) Instead of using molar concentrations, you can use gas pressures to calculate an equilibrium constant, in which case it is often labeled \(K_{\mathrm{p}}\) instead of \(K_{\mathrm{eq}} .\) For this reaction, \(K_{\mathrm{p}}=3.40 \mathrm{~atm}^{-1}\) at \(1000 \mathrm{~K}\). Consider a flask filled with all three of these gases such that \(\mathrm{p}_{\mathrm{SO} 2}=0.48 \mathrm{~atm}, p_{\mathrm{O} 2}=0.18 \mathrm{~atm}\), and \(\mathrm{p}_{\mathrm{SO} 3}=0.72 \mathrm{~atm} .\) Is this reaction at equilibrium or not? If not, which way will it shift to get there? (Hint: Write the equilibrium expression for \(K_{\mathrm{p}}\) in terms of the pressures, and then plug in the given values for the pressures. Then compare the value you get to the value of \(K_{\mathrm{p}}\) at this temperature.)

Step by step solution

01

Write the expression for Kp

For the given equilibrium reaction: \(2 SO_{2}(g) + O_{2}(g) \rightleftarrows 2 SO_{3}(g)\) We can write the expression for Kp as: \(K_p = \frac{[SO_3]^2}{[SO_2]^2[O_2]}\), where [SO3], [SO2], and [O2] are the equilibrium partial pressures of SO3, SO2, and O2, respectively.

02

Substitute the given pressure values

Now, we'll substitute the given pressure values into the Kp expression: \(K'_p = \frac{(0.72 \,atm)^2}{(0.48 \,atm)^2(0.18 \,atm)}\)

03

Calculate the value of Kp'

Calculate Kp': \(K'_p = \frac{(0.72 \,atm)^2}{(0.48 \,atm)^2(0.18 \,atm)} = \frac{0.5184 \,atm^2}{0.041472 \,atm^3} = 12.50\, atm^{-1}\)

04

Compare the calculated Kp' value with the given Kp value

We have found that: \(K'_p = 12.50 \,atm^{-1}\), while the given Kp at 1000 K is: \(K_p = 3.40 \,atm^{-1}\). As we can observe \(K'_p > K_p\), which means the reaction is not at equilibrium.

05

Determine the direction in which the reaction will shift

Since \(K'_p > K_p\), this indicates that the reaction will shift to the left to consume more SO3 and produce more SO2 and O2 to reach equilibrium. In conclusion, the reaction is not at equilibrium under the given conditions, and it will shift to the left (towards the reactants) to reach equilibrium.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant

In chemistry, the equilibrium constant (denoted as K or K_eq) is a number that expresses the ratio of the concentration of the products to the reactants at equilibrium, according to the balanced chemical equation. For reactions involving gases, we use partial pressures instead of concentrations and denote the equilibrium constant as K_p, as seen in the given exercise with sulfur dioxide and oxygen reacting to form sulfur trioxide.

It's crucial to know that the equilibrium constant is only affected by temperature changes and is constant for a given reaction at a specific temperature. In the exercise, the value of K_p is 3.40 atm^-1 at 1000 K. This doesn't indicate the extent of the reaction but rather the position of equilibrium under these conditions. When the system is at equilibrium, the forward and reverse reactions occur at the same rate, which gives us the magic K_p value unique to each reaction's condition.

Partial Pressure

Understanding Partial Pressure

Partial pressure refers to the pressure that each gas in a mixture would exert if it alone occupied the entire volume of the mixture at the same temperature. In the context of chemical equilibrium in gaseous systems, the equilibrium constant involving partial pressures, K_p, is used instead of concentrations. The partial pressures of SO₂, O₂, and SO₃ in the exercise represent how much each gas contributes to the total pressure of the system.

The equilibrium expression includes these partial pressures raised to the power of their respective coefficients in the balanced chemical equation. Accurate measurements of the partial pressures are necessary to determine the reaction quotient and to compare it with the equilibrium constant, which lets us predict the direction of the shift needed to reach equilibrium.

Le Chatelier's Principle

Shifting Equilibrium: Le Chatelier's Principle

Le Chatelier's Principle provides insight into the behavior of a chemical system at equilibrium when it is subjected to changes. According to this principle, if a dynamic equilibrium is disturbed by changing the conditions, such as concentration, pressure, volume, or temperature, the system adjusts itself to partially counteract the effect of the disturbance and restore a new equilibrium.

In the case of our exercise, when the calculated reaction quotient, K'_p, is greater than the equilibrium constant, K_p, it implies that there are too many products compared to what is expected at equilibrium. Thus, in compliance with Le Chatelier's Principle, this system will shift to the left, towards the reactants, to reduce the number of products and increase the number of reactants until the reaction quotient matches the K_p value again—that is, until equilibrium is re-established.

Reaction Quotient

The reaction quotient (Q) is a crucial concept used to predict the direction in which a reaction at non-equilibrium conditions will proceed to reach equilibrium. It is calculated using the same expression as the equilibrium constant (K), but with the initial concentrations or partial pressures. Comparing Q to K gives us information on the reaction's position: if Q < K, the reaction will proceed forward (right); if Q > K, it will move in the reverse direction (left).

From the exercise, by calculating K'_p (which is analogous to Q in this context), we determined that K'_p is significantly higher than K_p. This difference signifies that the reaction favors the reactants at the moment and that the system is not at equilibrium. Consequently, the reaction will shift toward the production of reactants (to the left) to attain equilibrium, in line with the guidelines of Le Chatelier's Principle.