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Chapter 13: Problem 28

At a particular temperature a 2.00-L flask at equilibrium contains \(2.80 \times 10^{-4} \mathrm{~mol} \mathrm{~N}_{2}, 2.50 \times 10^{-5} \mathrm{~mol} \mathrm{O}_{2}\), and \(2.00 \times 10^{-2}\) mol \(\mathrm{N}_{2} \mathrm{O} .\) Calculate \(K\) at this temperature for the reaction $$ 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{~N}_{2} \mathrm{O}(g) $$ If \(\left[\mathrm{N}_{2}\right]=2.00 \times 10^{-4} M,\left[\mathrm{~N}_{2} \mathrm{O}\right]=0.200 M\), and \(\left[\mathrm{O}_{2}\right]=0.00245\) \(M\), does this represent a system at equilibrium?

Short Answer

Expert verified
The equilibrium constant, K, for the given reaction at this temperature is approximately 64.20. When plugging in the provided concentrations, the reaction quotient, Q, is calculated to be approximately 19854.12. Since Q is not equal to K, the system is not at equilibrium.

Step by step solution

01

Write the equilibrium expression for K

First, write the equilibrium expression for the reaction: \(K = \frac{[\mathrm{N}_{2}\mathrm{O}]^2}{[\mathrm{N}_{2}]^2 \times [\mathrm{O}_{2}]}\)
02

Convert moles to molarity

Since we're given the amounts of substances in moles and the volume of the flask, we can find the molarity of each substance: Molarity of \(\mathrm{N}_{2} = \frac{2.80 \times 10^{-4}\,\mathrm{mol}}{2.00\,\mathrm{L}} = 1.40 \times 10^{-4}\,\mathrm{M}\) Molarity of \(\mathrm{O}_{2} = \frac{2.50 \times 10^{-5}\,\mathrm{mol}}{2.00\,\mathrm{L}} = 1.25 \times 10^{-5}\,\mathrm{M}\) Molarity of \(\mathrm{N}_{2}\mathrm{O} = \frac{2.00 \times 10^{-2}\,\mathrm{mol}}{2.00\,\mathrm{L}} = 1.00 \times 10^{-2}\,\mathrm{M}\)
03

Calculate K using the equilibrium concentrations

Plug the equilibrium concentrations into the expression for K: \(K = \frac{(1.00 \times 10^{-2})^2}{(1.40 \times 10^{-4})^2 \times (1.25 \times 10^{-5})} = 64.20\)
04

Plug the given concentrations into the reaction quotient, Q

For the given concentrations, \([\mathrm{N}_{2}] = 2.00 \times 10^{-4}\,\mathrm{M}\), \([\mathrm{N}_{2}\mathrm{O}] = 0.200 \,\mathrm{M}\), and \([\mathrm{O}_{2}] = 0.00245\,\mathrm{M}\): Reaction quotient, Q, is calculated the same way as K: \(Q = \frac{[\mathrm{N}_{2}\mathrm{O}]^{2}}{[\mathrm{N}_{2}]^{2}\times[\mathrm{O}_{2}]}\) Now, plug the given concentrations into the Q expression: \(Q = \frac{(0.200)^2}{(2.00 \times 10^{-4})^2 \times (0.00245)} = 19854.12\)
05

Compare Q and K to determine if the system is at equilibrium

The system is at equilibrium when Q = K. In this case, Q and K are not equal (Q > K), so the system is not at equilibrium.

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Most popular questions from this chapter

As we have attempted to lessen our dependence on fossil fuels, the demand for biofuels, such as ethanol, which is produced by the fermentation of the sugars found in corn, has increased. Using Le Châtelier's principle, predict which way the equilibrium will shift during the fermentation of sugar for each of the following changes. $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(a q) $$ a. when the concentration of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) is increased b. when the concentration of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) is decreased c. when \(\mathrm{CO}_{2}\) gas is added to the solution d. when the volume of water in the solution is doubled

Consider the reaction $$ \mathrm{P}_{4}(g) \longrightarrow 2 \mathrm{P}_{2}(g) $$ where \(K_{\mathrm{p}}=1.00 \times 10^{-1}\) at \(1325 \mathrm{~K}\). In an experiment where \(\mathrm{P}_{4}(g)\) is placed into a container at \(1325 \mathrm{~K}\), the equilibrium mixture of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g)\) has a total pressure of \(1.00 \mathrm{~atm} .\) Calculate the equilibrium pressures of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g) .\) Calculate the fraction (by moles) of \(\mathrm{P}_{4}(g)\) that has dissociated to reach equilibrium.

A 1.00-L flask was filled with \(2.00\) mol gaseous \(\mathrm{SO}_{2}\) and \(2.00 \mathrm{~mol}\) gaseous \(\mathrm{NO}_{2}\) and heated. After equilibrium was reached, it was found that \(1.30\) mol 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 (there may be more than one answer)? 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.

Novelty devices for predicting rain contain cobalt(II) chloride and are based on the following equilibrium: $$ \underset{\text { Purple }}{\mathrm{CoCl}_{2}(s)}+6 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \underset{\text { Pink }}{\mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(s)} $$ What color will such an indicator be if rain is imminent?
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