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At a particular temperature, \(8.0\) moles of \(\mathrm{NO}_{2}\) is placed into a 1.0-L container and the \(\mathrm{NO}_{2}\) dissociates by the reaction $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ At equilibrium the concentration of \(\mathrm{NO}(g)\) is \(2.0 M .\) Calculate \(K\) for this reaction.

Short Answer

Expert verified
The equilibrium constant, \(K\), for the given reaction is \(\frac{1}{9}\).

Step by step solution

01

Find the initial concentration of \(\mathrm{NO}_{2}\)

To find the initial concentration of \(\mathrm{NO}_{2}\), we use the formula: Concentration = moles / volume Here, we have 8.0 moles of \(\mathrm{NO}_{2}\) and the volume of the container is 1.0 L. So the initial concentration of \(\mathrm{NO}_{2}\) is: \[ \frac{8.0 \ \mathrm{moles}}{1.0 \ \mathrm{L}} = 8.0 \ \mathrm{M} \]
02

Find the change in concentration for each substance

Given the equilibrium concentration of \(\mathrm{NO}\), we can find the change in concentration of each substance during the reaction. Since the stoichiometry of the reaction is \(2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) + \mathrm{O}_{2}(g)\), for every 2 moles of \(\mathrm{NO}_{2}\) that dissociate, 2 moles of \(\mathrm{NO}\) and 1 mole of \(\mathrm{O}_{2}\) are produced. Let \(x\) be the change in concentration of \(\mathrm{NO}_{2}\) at equilibrium. The change in concentration of \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) at equilibrium can be expressed as follows: \(\Delta [\mathrm{NO}_{2}] = -2x\) \(\Delta [\mathrm{NO}] = +2x\) \(\Delta [\mathrm{O}_{2}] = +x\)
03

Calculate the concentrations at equilibrium

The concentrations of each substance at equilibrium are as follows: \([\mathrm{NO}_{2}]_{eq} = [\mathrm{NO}_{2}]_{0} + \Delta [\mathrm{NO}_{2}] = 8.0 - 2x\) \([\mathrm{NO}]_{eq} = [\mathrm{NO}]_{0} + \Delta [\mathrm{NO}] = 0 + 2x = 2.0 \ \mathrm{M}\) (given) \([\mathrm{O}_{2}]_{eq} = [\mathrm{O}_{2}]_{0} + \Delta [\mathrm{O}_{2}] = 0 + x\) Using the equilibrium concentration of \(\mathrm{NO}\), we can find the value of \(x\). \(2x = 2.0 \ \mathrm{M}\) \(x = 1.0 \ \mathrm{M}\) Now, we can find the equilibrium concentrations of \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\): \([\mathrm{NO}_{2}]_{eq} = 8.0 - 2(1.0) = 6.0 \ \mathrm{M}\) \([\mathrm{O}_{2}]_{eq} = 1.0 \ \mathrm{M}\)
04

Calculate the equilibrium constant, \(K\)

The equilibrium expression for the given reaction is: \(K = \frac{[\mathrm{NO}]^{2}[\mathrm{O}_{2}]}{[\mathrm{NO}_{2}]^{2}}\) Now, we can plug in the equilibrium concentrations: \(K = \frac{(2.0)^{2} \times 1.0}{(6.0)^{2}} = \frac{4}{36} = \frac{1}{9}\) So the equilibrium constant, \(K\), for this reaction is \(\frac{1}{9}\).

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Key Concepts

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

Chemical Equilibrium
Imagine a busy marketplace where products come in and go out at the same rate, maintaining a perfect balance. In chemistry, chemical equilibrium is a similar concept where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products over time. However, this doesn't mean the reactions have stopped—rather, they continue to occur simultaneously at equal rates.

For instance, in the reaction \(2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) + \mathrm{O}_{2}(g)\), when the system reaches equilibrium, the formation of nitric oxide (NO) and oxygen (O2) from nitrogen dioxide (NO2) is balanced by the reverse process converting NO and O2 back to NO2. Crucially, the equilibrium state does not signify equal concentrations but balanced reaction rates; significance lies in the fact that the system, once it reaches equilibrium, will have constant properties as long as external conditions do not change.
Reaction Stoichiometry
The art of cooking often involves precise measurements and ratios—its scientific counterpart in chemistry is reaction stoichiometry. This concept deals with the quantitative relationship between reactants and products in a chemical reaction. Stoichiometry allows chemists to predict the amounts of substances consumed and products formed in a reaction based on balanced chemical equations.

In the reaction under study, the stoichiometry indicates that two moles of NO2 will produce two moles of NO and one mole of O2. Understanding these ratios is critical not just for theoretical calculations but also for real-world applications, as it dictates the proportions of reactants necessary for a desired outcome. Hence, stoichiometry is a fundamental concept enabling us to grasp the implications of the equilibrium concentrations of each substance involved in a reaction.
Equilibrium Concentrations
Just as knowing the remaining slices of pizza can tell us how much everyone ate, in chemistry, equilibrium concentrations inform us about the extent of a reaction. These concentrations refer to the amounts of reactants and products present when a chemical reaction reaches equilibrium.

To calculate these concentrations, one often performs an 'ICE' analysis—an acronym for Initial, Change, and Equilibrium. This involves setting up a table to track the initial concentrations of reactants and products, the changes they undergo as the reaction proceeds, and their concentrations at equilibrium. In the given exercise, understanding the stoichiometry of the reaction and the initial concentration of NO2 allowed us to use the equilibrium concentration of NO to solve for the equilibrium concentrations of all species. This process involves algebraic calculations to determine the extent of the reaction, and thus the remaining concentrations of reactants, and in turn, the equilibrium constant, K, can give us an idea of the 'position' of equilibrium—favoring products, reactants, or a balanced state.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}, K=0.090\) for the reaction $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) $$ Calculate the concentrations of all species at equilibrium for each of the following cases. a. \(1.0 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\) and \(2.0 \mathrm{~g} \mathrm{Cl}_{2} \mathrm{O}\) are mixed in a \(1.0-\mathrm{L}\) flask. b. \(1.0\) mole of pure \(\mathrm{HOCl}\) is placed in a \(2.0-\mathrm{L}\) flask.

At a particular temperature, \(K=1.00 \times 10^{2}\) for the reaction $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ In an experiment, \(1.00\) mole of \(\mathrm{H}_{2}, 1.00 \mathrm{~mole}\) of \(\mathrm{I}_{2}\), and \(1.00\) mole of HI are introduced into a \(1.00-\mathrm{L}\) container. Calculate the concentrations of all species when equilibrium is reached.

Methanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: $$ \mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{H}_{2}(a q) $$ Assuming the value of \(K\) for this reaction is \(3.7 \times 10^{-10}\), what are the equilibrium concentrations of each species if you start with a \(1.24 M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

Consider the decomposition equilibrium for dinitrogen pentoxide: $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ At a certain temperature and a total pressure of \(1.00 \mathrm{~atm}\), the \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(0.50 \%\) decomposed (by moles) at equilibrium. a. If the volume is increased by a factor of \(10.0\), will the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposed at equilibrium be greater than, less than, or equal to \(0.50 \%\) ? Explain your answer. b. Calculate the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) that will be decomposed at equilibrium if the volume is increased by a factor of \(10.0 .\)

For the following endothermic reaction at equilibrium: $$ 2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) $$ which of the following changes will increase the value of \(K ?\) a. increasing the temperature b. decreasing the temperature c. removing \(\mathrm{SO}_{3}(g)\) (constant \(T\) ) d. decreasing the volume (constant \(T\) ) e. adding \(\operatorname{Ne}(g)\) (constant \(T\) ) f. adding \(\mathrm{SO}_{2}(g)\) (constant \(T\) ) g. adding a catalyst (constant \(T\) )

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