The equilibrium constant for the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g})\) has the value \(K=2.5 \times 10^{10}\) at \(500 \mathrm{~K}\). Find the value of \(K\) for each of the following reactions at the same temperature. (a) \(\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})=\mathrm{SO}_{3}(\mathrm{~g})\) (b) \(\mathrm{SO}_{3}(\mathrm{~g})=\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) (c) \(3 \mathrm{SO}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 3 \mathrm{SO}_{3}(\mathrm{~g})\)

Short Answer

Expert verified
The value of K for (a) is approximately \(5 \times 10^5\), for (b) approximately \(4 \times 10^{-11}\), and for (c) approximately \(1.5625 \times 10^{31}\).

Step by step solution

01

Understanding Equilibrium Constant Relations

The equilibrium constant, K, for a reaction is related to the stochiometry of the reaction. When a balanced chemical equation is multiplied by a coefficient, c, the equilibrium constant for the new equation is K^c. When a reaction is reversed, the new equilibrium constant is the reciprocal of the original equilibrium constant, 1/K.
02

Calculate K for Reaction (a)

Reaction (a) is half the equation of the given reaction. Therefore, since the coefficients are halved, we take the square root of the original K to get the new K: \( K' = \sqrt{K} = \sqrt{2.5 \times 10^{10}} \).
03

Calculate K for Reaction (b)

Reaction (b) is the reverse of the given reaction. To find the new K, we take the reciprocal of the original K: \( K'' = \frac{1}{K} = \frac{1}{2.5 \times 10^{10}} \).
04

Calculate K for Reaction (c)

Reaction (c) is three times the equation of the given reaction. To find the new K, we raise the original K to the third power: \( K''' = K^3 = (2.5 \times 10^{10})^3 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. This means that the concentrations of reactants and products remain constant over time, though they are not necessarily equal. In this balanced state, no net change in the quantity of reactants and products occurs.

Equilibrium does not indicate that the reactants and products are in equal concentrations, but rather that their ratios are constant. The equilibrium constant, represented as K, is a numerical value that expresses this ratio at a given temperature. Specifically, for the reaction at hand, 2 SO_2(g) + O_2(g) = 2 SO_3(g), K represents the concentrations of the products (SO_3) raised to their stoichiometric coefficients divided by the concentrations of the reactants (SO_2 and O_2), each raised to their respective stoichiometric coefficients.
Reaction Quotient
The reaction quotient, Q, is a measure that tells us how close a system is to reaching equilibrium. It is calculated in the same way as the equilibrium constant, using the current concentrations of the products and reactants.

If Q is less than K, the forward reaction is favored, and the concentrations of products will increase until equilibrium is reached. If Q is greater than K, the reverse reaction is favored, and the concentrations of reactants will increase until equilibrium is achieved. When Q equals K, the system is at equilibrium.
Le Chatelier's Principle
Le Chatelier's principle is crucial for understanding how a system at equilibrium responds to changes in concentration, pressure, or temperature. Essentially, it states that if an external stress is applied to a system at equilibrium, the system will adjust itself to minimize the impact of that change.

For instance, increasing the concentration of reactants will cause the equilibrium to shift towards forming more products to reduce the reactant concentration, thus favoring the forward reaction. Conversely, decreasing the pressure by increasing volume will shift the equilibrium towards the side with more gas molecules. Adjusting temperature can also affect the equilibrium; for exothermic reactions, increasing temperature will shift the equilibrium towards the reactants, while for endothermic reactions, it will shift towards the products.
Gas Phase Reactions
Gas phase reactions are chemical processes that occur between substances in the gaseous state. One characteristic of gas phase reactions is that changes in pressure and volume can significantly affect the point of equilibrium, according to Le Chatelier's principle.

For the given gas phase reaction involving sulfur dioxide and oxygen forming sulfur trioxide, the reaction's equilibrium can be influenced by such changes. As it's a balanced equation with an equal number of gas molecules on both the reactant and product sides, changing the volume or pressure would typically have no net effect on the direction of the shift in equilibrium. However, for gas-phase reactions where there are different numbers of gas molecules on each side, the effects can be quite pronounced.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A mixture consisting of \(1.000 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) and \(1.000 \mathrm{~mol} \mathrm{CO}(\mathrm{g})\) is placed in a \(10.00-\mathrm{L}\) reaction vessel at \(800 \mathrm{~K}\). At equilibrium, \(0.665 \mathrm{~mol} \mathrm{CO}_{2}(\mathrm{~g})\) is present as a result of the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})=\) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\). What are (a) the equilibrium concentrations for all substances and (b) the value of \(K_{e}\) ?

Explain why the following equilibria are heterogencous and write the reaction quoticnt \(Q\) for each one. (a) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})=\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{HCl}(\mathrm{g})\) (b) \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{OH}_{2} \mathrm{O}\) (s) \(\rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{~s})+10 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(2 \mathrm{KCO}_{3}\) (s) \(\Rightarrow 2 \mathrm{KCl}\) (s) \(+3 \mathrm{O}_{2}\) (g)

(a) Calculate the reaction free energy of \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Hl}(\mathrm{g})\) at \(700 \mathrm{~K}\) when the concentrations of \(\mathrm{H}_{2}, \mathrm{I}_{2}\), and HI are \(0.026,0.33\), and \(1.84 \mathrm{~mol} \cdot \mathrm{L}^{-1}\), respectively. For this reaction, \(K_{c}=54\) at \(700 \mathrm{~K}\). (b) Indicate whether this reaction mixture is likely to form reactants, is likely to form products, or is at equilibrium.

If \(Q=1.0\) for the reaction \(N_{2}(g)+O_{2}(g) \rightarrow\) \(2 \mathrm{NO}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\), will the reaction have a tendency to form products or reactants, or will it be at equilibrium?

Analysis of a reaction mixture showed that it had the composition \(0.417 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}\), \(0.524 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\), and \(0.122 \mathrm{~mol}-\mathrm{L}^{-1} \mathrm{NH}_{3}\) at \(800 \mathrm{~K}\), at which temperature \(K_{c}=0.278\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})=2 \mathrm{NH}_{3}(\mathrm{~g}) \cdot(\mathrm{a}) \mathrm{Calculare}\) the reaction quotient. (b) Is the reaction mixture at equilibrium? (c) If not, is there a tendency to form more reactants or more products?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free