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Chapter 13: Q74E (page 759)

A student solved the following problem and found \(\left[ {{N_2}{O_4}} \right] = 0.16M\)at equilibrium. How could this student recognize that the answer was wrong without reworking the problem? The problem was: What is the equilibrium concentration of \(\left[ {{N_2}{O_4}} \right]\) in a mixture formed from a sample of \(N{O_2}\) with a concentration of \(0.10M\)?

\(2N{O_2}(g) \rightleftharpoons {N_2}{O_4}(g)\)

\({K_c} = 160\)

Short Answer

Expert verified

\(\begin{array}{c}\left[ {{{\rm{N}}_2}{{\rm{O}}_4}} \right] = 0.0419{\rm{M}}\\\left[ {{\rm{N}}{{\rm{O}}_2}} \right] = 0.0162M\end{array}\)

The value obtained for \({{\rm{N}}_2}{{\rm{O}}_4}\) does not matches with the student鈥檚 result. So, the student鈥檚 results were wrong

Step by step solution

01

Find out how the student’s result was wrong:

\(2N{O_2}(g) \rightleftharpoons {N_2}{O_4}(g)\)

Initial(M)

\(0.10\)

\(0\)

Change(M)

\( - 2x\)

\( + x\)

Equilibrium(M)

\(0.10 - 2x\)

\(x\)

02

Find the value of x

\(\begin{array}{c}{K_C} = \frac{{\left[ {{{\rm{N}}_2}{{\rm{O}}_4}} \right]}}{{{{\left[ {2{\rm{N}}{{\rm{O}}_2}} \right]}^2}}}\\160 = \frac{x}{{{{\left( {0.10 - 2x} \right)}^2}}}\\x = 160\left( {0.01 - 0.4x + 4{x^2}} \right)\\640{x^2} - 65x + 1.6 = 0\\x = 0.0419M\end{array}\)

Therefore, the equilibrium concentration will be

\(\begin{array}{c}\left[ {{{\rm{N}}_2}{{\rm{O}}_4}} \right] = x = 0.0419{\rm{M}}\\\left[ {{\rm{N}}{{\rm{O}}_2}} \right] = 0.10 - 2x = 0.0162M\end{array}\)

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Most popular questions from this chapter

Question: What is the minimum mass of CaCO3 required to establish equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant (Kc) is 0.050 for the decomposition reaction of CaCO3 at that temperature?

Explain how to recognize the conditions under which changes in pressure would affect systems at equilibrium.

What are the concentrations of \(PC{l_5}\), \(PC{l_3}\), and \(C{l_2}\) in an equilibrium mixture produced by the decomposition of a sample of pure \(PC{l_5}\) with \([PC{l_5}] = 2.00M?\) \(PC{l_5}(g) \rightleftharpoons PC{l_3}(g) + {\mathbf{C}}{{\mathbf{l}}_{\mathbf{2}}}(g)\,\,\,\,\,\,\,\;{\mathbf{Kc}} = {\mathbf{0}}.{\mathbf{0}}211\)

Question: The binding of oxygen by hemoglobin (Hb), giving oxyhemoglobin (HbO2), is partially regulated by the concentration of H3O+ and dissolved CO2 in the blood. Although the equilibrium is complicated, it can be summarized as

HbO2(aq) + H3 O+(aq) + CO2(g) 鈬� CO2 鈭扝产鈭扝+ + O2(g) + H2 O(l)

(a) Write the equilibrium constant expression for this reaction.

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