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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’s result. So, the student’s 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

What is the approximate value of the equilibrium constant KP for the change C2H5OC2H5 (l)⇌C2 H5OC2H5 (g) at 25 °C.

(Vapor pressure was described in the previous chapter on liquids and solids; refer back to this chapter to find the relevant information needed to solve this problem.)

Question : A 0.010Msolution of the weak acid HA has an osmotic pressure (see chapter on solutions and colloids) of 0.293 atm at 25 °C. A 0.010Msolution of the weak acid HB has an osmotic pressure of 0.345 atm under the same conditions.

(a) Which acid has the larger equilibrium constant for ionization

HA[HA(aq) ⇌ A−(aq) + H+(aq)]or HB[HB(aq) ⇌ H+(aq) + B−(aq)]?

(b) What are the equilibrium constants for the ionization of these acids?

(Hint: Remember that each solution contains three dissolved species: the weak acid (HA or HB), the conjugate base (A− or B−), and the hydrogen ion (H+). Remember that osmotic pressure (like all colligative properties) is related to the total number of solute particles. Specifically for osmotic pressure, those concentrations are described by molarities.)

Question: The hydrolysis of the sugar sucrose to the sugars glucose and fructose follows a first-order rate equation for the disappearance of sucrose.

C12 H22 O11(aq) + H2O(l)⟶C6 H12 O6 (aq) + C6 H12 O6 (aq)

Rate = k[C12H22O11]

In neutral solution, k = 2.1 × 10−11/s at 27 °C. (As indicated by the rate constant, this is a very slow reaction. In the human body, the rate of this reaction is sped up by a type of catalyst called an enzyme.) (Note: That is not a mistake in the equation—the products of the reaction, glucose and fructose, have the same molecular formulas, C6H12O6, but differ in the arrangement of the atoms in their molecules). The equilibrium constant for the reaction is 1.36 × 105 at 27 °C. What are the concentrations of glucose, fructose, and sucrose after a 0.150 M aqueous solution of sucrose has reached equilibrium? Remember that the activity of a solvent (the effective concentration) is 1.

Carbon reacts with water vapor at elevated temperatures

\(C(s) + {H_2}O(g) \rightleftharpoons CO(g) + {H_2}(g)\)

\({K_c} = 0.2at100{0^o}C\)

What is the concentration of CO in an equilibrium mixture with \(\left[ {{H_2}O} \right] = 0.500M\)at \(100{0^0}C\)?

The following reaction has \({K_P} = 4.50 \times {10^{ - 5}}\) at \(720\;{\rm{K}}\).

\({{\rm{N}}_2}(g) + 3{{\rm{H}}_2}(g) \rightleftharpoons 2{\rm{N}}{{\rm{H}}_3}(g)\)

If a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium?

\(P\left( {{\rm{N}}{{\rm{H}}_3}} \right) = 93\;{\rm{atm}},\;P\left( {\;{{\rm{N}}_2}} \right) = 48\;{\rm{atm}},\;{\rm{and}}\;P\left( {{{\rm{H}}_2}} \right) = 52\)
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