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Chapter 13: Q78E (page 760)

What are all concentrations after a mixture that contains \(\left[ {{{\bf{H}}_{\bf{2}}}{\bf{O}}} \right] = {\bf{1}}.{\bf{00Mand}}\left[ {{\bf{C}}{{\bf{l}}_{\bf{2}}}{\bf{O}}} \right] = {\bf{1}}.{\bf{00M}}\) comes to equilibrium at \({\bf{25}}^\circ {\bf{C}}\)?

\({{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}(g) + {\mathbf{C}}{{\mathbf{l}}_{\mathbf{2}}}{\mathbf{O}}(g) \rightleftharpoons {\mathbf{2HOCl}}(g);\;{\mathbf{Kc}} = {\mathbf{0}}.{\mathbf{0900}}\)

Short Answer

Expert verified

Equilibrium concentration of all species are:

\(\begin{array}{*{20}{c}}{[HOCl] = 2x = 0.26M}\\{\,\,\,\,\,\,\,\left[ {{H_2}O} \right] = 1 - x = 0.87M}\\{\,\,\,\,\,\,\left[ {C{l_2}O} \right] = 1 - x = 0.87M}\end{array}\)

Step by step solution

01

Define concentration.

It is a constituent divided by total volume of a mixture.

02

Find equilibrium concentration

Let the change in concentration be x

Therefore, equilibrium concentration of all the species will be

\(\begin{array}{*{20}{c}}{[HOCl] = 2x}\\{\,\,\,\,\,\,\left[ {{H_2}O} \right] = 1 - x}\\{\,\,\,\,\,\left[ {C{l_2}O} \right] = 1 - x}\end{array}\)

Equilibrium constant can be written as:

\(\begin{array}{*{20}{c}}{\,\,\,\,\,\,\,\,\,{K_c} = \frac{{{{[HOCl]}^2}}}{{\left[ {{H_2}O} \right] \times \left[ {C{l_2}O} \right]}}}\\{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.0900 = \frac{{{{(2x)}^2}}}{{(1.00 - x) \times (1.00 - x)}}}\\{0.0900 = \frac{{4{x^2}}}{{1.00 - 2x + {x^2}}}}\end{array}\)

\(\begin{array}{*{20}{c}}{4{x^2} = 0.09 - 0.18x + 0.09{x^2}}\\{\,\,\,\,\,0 = 3.91{x^2} + 0.18x - 0.09}\\{{\rm{\;Using equation solver, we get\;}}}\\{x = 0.13{\rm{M}}}\end{array}\)

Therefore, equilibrium concentrations are:

\(\begin{array}{*{20}{c}}{[{\rm{HOCl}}] = 2{\rm{x}} = 0.26{\rm{M}}}\\{\,\,\,\,\,\,\left[ {{{\rm{H}}_2}{\rm{O}}} \right] = 1 - {\rm{x}} = 0.87{\rm{M}}}\\{\,\,\,\,\left[ {{\rm{C}}{{\rm{l}}_2}{\rm{O}}} \right] = 1 - {\rm{x}} = 0.87{\rm{M}}}\end{array}\)

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Most popular questions from this chapter

Convert the values of Kc to values of KP or the values of KP to values of Kc .

\((a)\,C{l_2}\left( g \right) + B{r_2}\left( g \right)\rightleftharpoons 2BrCl(g)\,\,\,{K_C} = 4.7 \times {10^{ - 2}}\,at\,25^\circ C\)

\((b)2{\rm{S}}{{\rm{O}}_2}(g) + {{\rm{O}}_2}(g)\rightleftharpoons 2{\rm{S}}{{\rm{O}}_3}(g){K_P} = 48.2 at {500^\circ} {\rm{C}}\)

\((c){\rm{CaC}}{{\rm{l}}_2} \cdot 6{{\rm{H}}_2}{\rm{O}}(s)\rightleftharpoons {\rm{CaC}}{{\rm{l}}_2}(s) + 6{{\rm{H}}_2}{\rm{O}}(g){K_P} = 5.09 \times {10^{ - 44}}at{25^\circ }{\rm{C}}\)

\((d){{\rm{H}}_2}{\rm{O}}(l)\rightleftharpoons {{\rm{H}}_2}{\rm{O}}(g)\,{K_P} = 0.196\,at\,{60^\circ }{\rm{C}}\)

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) + H2翱(濒)鉄禖6 H12 O6 (aq) + C6 H12 O6 (aq)

Rate = k[C12H22O11]

In neutral solution, k = 2.1 脳 10鈭�11/s at 27 掳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鈥攖he 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 脳 105 at 27 掳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.

How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?

a. \(2{\rm{N}}{{\rm{H}}_3}(g)\rightleftharpoons {{\rm{N}}_2}(g) + 3{{\rm{H}}_2}(g)\) \({\rm{\Delta }}H = 92{\rm{kJ}}\)

b. \({{\rm{N}}_2}(g) + {{\rm{O}}_2}(g)\rightleftharpoons 2{\rm{NO}}(g)\) \({\rm{\Delta }}H = 181{\rm{kJ}}\)

c. \(2{{\rm{O}}_3}(g)\rightleftharpoons 3{{\rm{O}}_2}(g)\) \({\rm{\Delta }}H = - 285{\rm{kJ}}\)

d.\({\rm{CaO(s) + C}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}}\rightleftharpoons {\rm{CaC}}{{\rm{O}}_{\rm{3}}}{\rm{(s)}}\) \({\rm{\Delta }}H = - 176{\rm{kJ}}\)

When heated, iodine vapor dissociates according to this equation: I2 (g) 鈬� 2I (g). At 1274K a sample exhibits a partial pressure of I2 of 0.1122 and a partial pressure due to I atoms of 0.1378 atm. Determine the value of the equilibrium constant, Kp for the decomposition at 1274K

Question: A 1.00-L vessel at 400 掳C contains the following equilibrium concentrations: N2, 1.00M; H2, 0.50M; and NH3, 0.25M. How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1M?
See all solutions

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