A 1.00-L flask was filled with \(2.00\) moles of gaseous \(\mathrm{SO}_{2}\) and \(2.00\) moles of gaseous \(\mathrm{NO}_{2}\) and heated. After equilibrium was reached, it was found that \(1.30\) moles of gaseous NO was present. Assume that the reaction $$ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g) $$ occurs under these conditions. Calculate the value of the equilibrium constant, \(K\), for this reaction.

Short Answer

Expert verified
The equilibrium constant (K) for the given reaction is 3.4 under the specified conditions.

Step by step solution

01

Write the balanced chemical equation #

We are given the balanced chemical equation as: \[\mathrm{SO_{2}(g) + NO_{2}(g) \rightleftharpoons SO_{3}(g) + NO(g)}\]
02

Determine the initial moles for each species #

We are given the initial number of moles as: SO2: 2.00 moles NO2: 2.00 moles Since the reaction has not started yet, the initial moles of SO3 and NO are zero.
03

Analyze the moles of NO at equilibrium #

We are given that at equilibrium, there are 1.30 moles of NO present. When NO is formed in the reaction, an equal number of moles of NO2 and SO2 react. So, the change in moles of SO2 and NO2 is equal to the moles of NO formed (1.30 moles). The change in moles of SO3 is also equal to the moles of NO formed. Thus, the number of moles of each species at equilibrium is: SO2: 2.00 - 1.30 = 0.70 moles NO2: 2.00 - 1.30 = 0.70 moles SO3: 0 + 1.30 = 1.30 moles NO: 1.30 moles
04

Calculate the concentration of each gas at equilibrium #

As the volume of the flask is 1.00 L, the concentration of each gas can be calculated by dividing the number of moles by the volume: \[ [\mathrm{SO_{2}}] = \frac{0.70 \, \text{moles}}{1.00 \, \text{L}} = 0.70 \, \mathrm{M} \] \[ [\mathrm{NO_{2}}] = \frac{0.70 \, \text{moles}}{1.00 \, \text{L}} = 0.70 \, \mathrm{M} \] \[ [\mathrm{SO_{3}}] = \frac{1.30 \, \text{moles}}{1.00 \, \text{L}} = 1.30 \, \mathrm{M} \] \[ [\mathrm{NO}] = \frac{1.30 \, \text{moles}}{1.00 \, \text{L}} = 1.30 \, \mathrm{M} \]
05

Write the expression for K and calculate its value #

The equilibrium constant (K) can be expressed as: \[K = \frac{[\mathrm{SO_{3}}][\mathrm{NO}]}{[\mathrm{SO_{2}}][\mathrm{NO_{2}}]}\] Now plug in the values for the concentrations: \[K = \frac{(1.30)(1.30)}{(0.70)(0.70)} = \frac{1.69}{0.49} = 3.4\] The value of the equilibrium constant, K, for the given reaction is 3.4 under the specified conditions.

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is crucial for interpreting how reactions behave in closed systems. At the core, chemical equilibrium is a state in which the rate of the forward reaction equals the rate of the reverse reaction; the concentrations of reactants and products remain constant over time, but not necessarily equal. It's an essential concept, especially when considering reversible reactions, like the synthesis of SO3 and NO from SO2 and NO2.

In our exercise, once the system reaches equilibrium, the amount of SO2 and NO2 gases transformed into SO3 and NO no longer changes. This doesn't mean the reaction has stopped—indeed, reactions continue to occur in both directions, but at an equal rate, maintaining a balance of the reactant and product concentrations.
Reaction Quotient
The reaction quotient, Q, is a measure that tells us the direction in which a reaction will proceed to reach equilibrium. It is calculated using the same formula as the equilibrium constant (K), but with the current concentrations, not necessarily at equilibrium. For the reaction between SO2 and NO2 forming SO3 and NO, you would calculate the reaction quotient similarly to how you would find K.

If Q is less than K, the forward reaction is favored, and the system will produce more products. If Q is greater than K, the system will form more reactants. If Q equals K, the system is at equilibrium. This principle assists in predicting the behavior of the system upon changes in concentration, pressure, or volume.
Moles and Molarity
The relationship between moles and molarity is a fundamental aspect of chemistry, especially when dealing with reactions in solution. The number of moles refers to the quantity of substance, while molarity (M) is a measure of concentration, specifically, moles per liter of solution. In the given exercise, the molarity of each gas at equilibrium is derived by dividing the number of moles by the volume of the flask.

For instance, if you have 0.70 moles of SO2 in a 1.00 L flask, the molarity is 0.70 M. Understanding this relationship allows for the accurate calculation of the concentrations needed to determine the equilibrium constant, which is vital for predicting the extent and direction of a chemical reaction.
Le Chatelier's Principle
Le Chatelier's Principle predicts how a change in conditions can affect the position of equilibrium in a chemical reaction. According to this principle, if a system at equilibrium experiences a change in concentration, temperature, or pressure, the system will adjust to counteract that change and re-establish equilibrium.

In the context of the provided exercise, if more SO2 were added to our system, Le Chatelier's Principle suggests the equilibrium would shift to the right, forming more SO3 and NO, to reduce the effect of the added SO2. Conversely, if the temperature changed, it would affect the equilibrium position based on the exothermic or endothermic nature of the reaction. This principle is invaluable for fine-tuning reactions in industrial processes and predicting how systems respond to external stress.

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Most popular questions from this chapter

The equilibrium constant \(K_{\mathrm{p}}\) for the reaction $$ \mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g) $$ at \(700^{\circ} \mathrm{C}\) is \(0.76 .\) Detemine the initial pressure of carbon tetrachloride that will produce a total equilibrium pressure of \(1.20 \mathrm{~atm}\) at \(700^{\circ} \mathrm{C}\).

In which direction will the position of the equilibrium $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ be shifted for each of the following changes? a. \(\mathrm{H}_{2}(g)\) is added. b. \(\mathrm{I}_{2}(g)\) is removed. c. \(\mathrm{HI}(g)\) is removed. d. In a rigid reaction container, some \(\operatorname{Ar}(g)\) is added. e. The volume of the container is doubled. f. The temperature is decreased (the reaction is exothermic).

Le Châtelier's principle is stated (Section \(13.7)\) as follows: "If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change." The system \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in \(\mathrm{H}_{2}\) concentration and an increase in \(\mathrm{NH}_{3}\) concentration. In the experiment the volume is assumed to be constant. On the other hand, if \(\mathrm{N}_{2}\) is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount of \(\mathrm{NH}_{3}\) actually could decrease and the concentration of \(\mathrm{H}_{2}\) would increase as equilibrium is reestablished. Explain how this can happen. Also, if you consider this same system at equilibrium, the addition of an inert gas, holding the pressure constant, does affect the equilibrium position. Explain why the addition of an inert gas to this system in a rigid container does not affect the equilibrium position.

Predict the shift in the equilibrium position that will occur for each of the following reactions when the volume of the reaction container is increased. a. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) b. \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) d. \(\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)\) e. \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)

Lexan is a plastic used to make compact discs, eyeglass lenses, and bullet- proof glass. One of the compounds used to make Lexan is phosgene \(\left(\mathrm{COCl}_{2}\right)\), an extremely poisonous gas. Phosgene decomposes by the reaction $$ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ for which \(K_{\mathrm{p}}=6.8 \times 10^{-9}\) at \(100^{\circ} \mathrm{C}\). If pure phosgene at an jnitial pressure of \(1.0\) atm decomposes, calculate the equilibrium pressures of all species.

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