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At a particular temperature, 8.1 moles of \(\mathrm{NO}_{2}\) gas are placed in a 3.0 -L container. Over time the \(\mathrm{NO}_{2}\) decomposes to \(\mathrm{NO}\) and \(\mathrm{O}_{2} :\) $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the concentration of \(\mathrm{NO}(g)\) was found to be 1.4 \(\mathrm{mol} / \mathrm{L}\) . Calculate the value of \(K\) for this reaction.

Short Answer

Expert verified
The equilibrium constant, \(K\), for this reaction at the given temperature is approximately 0.939.

Step by step solution

01

Calculating initial concentrations

First, let's calculate the initial concentration of \(\mathrm{NO}_{2}\) by dividing the given moles (8.1 mol) by the volume of the container (3.0 L): \[\mathrm{NO}_{2} \text{ initial concentration} = \frac{8.1 \text{ moles}}{3.0 \text{ L}} = 2.7 \frac{\text{mol}}{\text{L}}\]
02

Setting up changes in concentration

Let's use the stoichiometric coefficients in the balanced chemical equation to understand the changes in concentrations during the reaction. At equilibrium, let the decrease in concentration of \(\mathrm{NO}_{2}\) be \(x\). Since two moles of \(\mathrm{NO}_{2}\) are required to produce two moles of \(\mathrm{NO}\) and one mole of \(\mathrm{O}_{2}\), the concentration of \(\mathrm{NO}_{2}\) will decrease by \(x\) molar, while the concentration of \(\mathrm{NO}\) will increase by \(x\) molar, and the concentration of \(\mathrm{O}_{2}\) will increase by \(\frac{x}{2}\) molar.
03

Calculating x

Given the equilibrium concentration of \(\mathrm{NO}(g)\) as 1.4 M, we can find the value of \(x\): \(x = \text{equilibrium concentration of } \mathrm{NO} = 1.4 \frac{\text{mol}}{\text{L}}\)
04

Finding equilibrium concentrations

Now that we have found the value of \(x\), we can find the equilibrium concentrations for all reactants and products: \[\text{Equilibrium concentration of } \mathrm{NO}_{2} = 2.7 \frac{\text{mol}}{\text{L}} - x = 2.7 - 1.4 = 1.3 \frac{\text{mol}}{\text{L}}\] \[\text{Equilibrium concentration of } \mathrm{NO} = x = 1.4 \frac{\text{mol}}{\text{L}}\] \[\text{Equilibrium concentration of } \mathrm{O}_{2} = \frac{x}{2} = \frac{1.4}{2} = 0.7 \frac{\text{mol}}{\text{L}}\]
05

Calculating the equilibrium constant, K

Using the equilibrium concentrations, we can write the expression for the equilibrium constant, \(K\): \[K = \frac{[\mathrm{NO}]^{2}[\mathrm{O}_{2}]}{[\mathrm{NO}_{2}]^{2}}\] Now, substitute the equilibrium concentrations into the expression: \[K = \frac{(1.4)^{2}(0.7)}{(1.3)^{2}} = \frac{1.586}{1.69}\] After performing the calculation: \[K \approx 0.939\]
06

Final Answer

The equilibrium constant, \(K\), for this reaction at the given temperature is approximately 0.939.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium describes the state where the concentrations of reactants and products remain constant over time in a closed system. This happens when the forward and reverse reactions occur at the same rate. In our reaction \( 2 \mathrm{NO}_{2} \rightleftharpoons 2 \mathrm{NO} + \mathrm{O}_{2} \), equilibrium is reached when the rate at which \( \mathrm{NO}_{2} \) decomposes matches the rate at which \( \mathrm{NO} \) and \( \mathrm{O}_{2} \) combine to reform \( \mathrm{NO}_{2} \).

At equilibrium, the concentrations don’t change, but that doesn’t mean they are equal. Instead, the relative concentrations help us determine the equilibrium constant \( K \), a powerful tool in predicting how a system will react to changes.
Reaction Stoichiometry
Reaction stoichiometry involves using the balanced chemical equation to relate the amounts of reactants and products. It's essential in understanding how changes in one part of a reaction affect the rest.

In this case, the stoichiometry of the reaction \( 2 \mathrm{NO}_{2} \rightarrow 2 \mathrm{NO} + \mathrm{O}_{2} \) tells us that two moles of \( \mathrm{NO}_{2} \) produce two moles of \( \mathrm{NO} \) and one mole of \( \mathrm{O}_{2} \). This allows us to calculate the changes in concentration as the reaction progresses.
  • \( \mathrm{NO}_{2} \) concentration decreases by \( x \) moles per liter.
  • \( \mathrm{NO} \) concentration increases by \( x \) moles per liter.
  • \( \mathrm{O}_{2} \) concentration increases by \( \frac{x}{2} \) moles per liter.
Understanding stoichiometry helps ensure that these calculations are precise, ensuring we accurately calculate equilibrium concentrations.
Concentration Calculations
Concentration calculations are crucial for finding the equilibrium constant. We begin by determining the initial concentration of \( \mathrm{NO}_{2} \). We do this by dividing the number of moles by the volume:

\[ \mathrm{NO}_{2} \text{ initial concentration} = \frac{8.1 \text{ moles}}{3.0 \text{ L}} = 2.7 \text{ mol/L} \]

Next, we calculate the equilibrium concentrations using given data. For \( \mathrm{NO}(g) \), the equilibrium concentration is provided as 1.4 mol/L. With this information, we can calculate the other concentrations:
  • \[ \text{\( \mathrm{NO}_{2} \) equilibrium concentration} = 2.7 - 1.4 = 1.3 \text{ mol/L} \]
  • \[ \text{\( \mathrm{O}_{2} \) equilibrium concentration} = \frac{1.4}{2} = 0.7 \text{ mol/L} \]
These calculations lay the foundation for finding the equilibrium constant, \( K \).
Gas Laws
Gas laws might not seem directly connected, but they help us understand reactions involving gases. In this exercise, knowing the relationships between pressure, volume, and temperature can enable us to predict how reaction rates might change with different conditions.

Although specific gas laws like Boyle’s Law or Charles' Law aren't directly used in the calculation of \( K \), understanding that gas concentrations can change with pressure and temperature is crucial in practical applications.

This insight allows chemists to manipulate conditions to favor the production of desired products, which is a valuable tool in industries such as pharmaceuticals and manufacturing.

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Most popular questions from this chapter

Consider the following reaction at a certain temperature: $$4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)$$ An equilibrium mixture contains 1.0 mole of \(\mathrm{Fe},\) \(1.0 \times 10^{-3}\) mole of \(\mathrm{O}_{2},\) and 2.0 \(\mathrm{moles}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) all in a 2.0 \(\mathrm{-L}\) container. Calculate the value of \(K\) for this reaction.

For the reaction $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2 \mathrm{CO}(g)$$ \(K_{\mathrm{p}}=2.00\) at some temperature. If this reaction at equilibrium has a total pressure of 6.00 \(\mathrm{atm}\) , determine the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the reaction container.

An important reaction in the commercial production of hydrogen is $$\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$$ How will this system at equilibrium shift in each of the five following cases? a. Gaseous carbon dioxide is removed. b. Water vapor is added. c. In a rigid reaction container, the pressure is increased by adding helium gas. d. The temperature is increased (the reaction is exothermic). e. The pressure is increased by decreasing the volume of the reaction container.

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.

The value of the equilibrium constant \(K\) depends on which of the following (more than one answer may be correct)? a. the initial concentrations of the reactants b. the initial concentrations of the products c. the temperature of the system d. the nature of the reactants and products Explain.

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