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Chapter 15: Problem 30

Is a mixture of \(0.0205 \mathrm{mol} \mathrm{NO}_{2}(\mathrm{g})\) and \(0.750 \mathrm{mol}\) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})\) in a \(5.25 \mathrm{L}\) flask at \(25^{\circ} \mathrm{C},\) at equilibrium? If not, in which direction will the reaction proceed toward products or reactants? $$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=4.61 \times 10^{-3} \mathrm{at} 25^{\circ} \mathrm{C}$$

Short Answer

Expert verified
No, the system is not at equilibrium. The reaction will proceed in the forward direction, towards the products.

Step by step solution

01

Calculate the initial concentrations

Calculate the molar concentration of both \(NO_2\) and \(N_2O_4\) at the onset using the given quantity and volume in the flask. The concentration (in Molarity, M) is calculated by dividing the number of moles by the volume of the solution in liters. The initial concentration of \(NO_2\) is \(0.0205 mol / 5.25 L = 0.00390 M\) and for \(N_2O_4\) is \(0.750 mol / 5.25 L = 0.143 M\)
02

Calculate the reaction quotient \(Q_c\)

The reaction quotient \(Q_c\), similar to \(K_c\), gives the measure of the ratio of the concentrations of the products raised to their stoichiometric coefficient to that of the reactants raised to their stoichiometric coefficients, at any point in time. Here, \(Q_c\) is calculated via: \(Q_c = [NO_2]^2 / [N_2O_4]\). The efficent concentrations are used. Thus \(Q_c = (0.00390)^2 / 0.143 = 1.065x10^-4\)
03

Compare \(Q_c\) to \(K_c\)

Now, compare \(Q_c\) to \(K_c\). If \(Q_c < K_c\), this means the system is not at equilibrium and the reaction will proceed in the forward direction to reach equilibrium, converting more reactants to products. If \(Q_c > K_c\), this means the system is not at equilibrium and the reaction will proceed in the reverse direction to reach equilibrium, converting more products to reactants. Here, \(1.065x10^-4 < 4.61x10^-3\), so the reaction will proceed toward the products side.

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Most popular questions from this chapter

The following data are given at \(1000 \mathrm{K}: \mathrm{CO}(\mathrm{g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) ; \quad \Delta H^{\circ}=-42 \mathrm{k} \mathrm{J}\) \(K_{\mathrm{c}}=0.66 .\) After an initial equilibrium is established in a \(1.00 \mathrm{L}\) container, the equilibrium amount of \(\mathrm{H}_{2}\) can be increased by (a) adding a catalyst; (b) increasing the temperature; (c) transferring the mixture to a 10.0 L container; (d) in some way other than (a), (b), Or (c).

An important environmental and physiological reaction is the formation of carbonic acid, \(\mathrm{H}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) from carbon dioxide and water. Write the equilibrium constant expression for this reaction in terms of activities. Convert that expression into an equilibrium constant expression containing concentrations and pressures.

Using the method in Appendix \(\mathrm{E}\), construct a concept map of Section \(15-6,\) illustrating the shift in equilibrium caused by the various types of disturbances discussed in that section.

Based on these descriptions, write a balanced equation and the corresponding \(K_{\mathrm{p}}\) expression for each reversible reaction. (a) Oxygen gas oxidizes gaseous ammonia to gaseous nitrogen and water vapor. (b) Hydrogen gas reduces gaseous nitrogen dioxide to gaseous ammonia and water vapor. (c) Nitrogen gas reacts with the solid sodium carbonate and carbon to produce solid sodium cyanide and carbon monoxide gas.

Determine \(K_{c}\) for the reaction \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})+\) \(\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g}),\) given the following data at \(298 \mathrm{K}\) $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=1.0 \times 10^{-9}$$ $$\operatorname{NOCl}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2} \mathrm{Cl}(\mathrm{g}) \quad K_{\mathrm{p}}=1.1 \times 10^{2}$$ $$\mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2} \mathrm{Cl}(\mathrm{g}) \quad K_{\mathrm{p}}=0.3$$
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