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Chapter 13: Q72E (page 759)

Calcium chloride 6鈭抙ydrate, \(CaC{l_2}.6{H_2}O\), dehydrates according to the equation

\(CaC{l_2} \times 6{H_2}O(s) \rightleftharpoons CaC{l_2}(s) + 6{H_2}O(g)\)

\({K_P} = 5.09 \times 1{0^{ - 44}}at2{5^o}C\)

What is the pressure of water vapor at equilibrium with a mixture of \(CaC{l_2}.6{H_2}O\)and \(CaC{l_2}\)?

Short Answer

Expert verified

The pressure of water vapor at equilibrium is \(6.09x1{0^{ - 8}}\)atm.

Step by step solution

01

Given information:

\(CaC{l_2}(s),6{H_2}O \rightleftharpoons CaC{l_2} + 6{H_2}O\)
  1. Value of equilibrium constant at \(2{5^0}C\)is \({K_P} = 5.09 \times 1{0^{ - 44}}\)

The value of equilibrium water vapor pressure needs to be calculated

02

Calculate the pressure of water vapor at equilibrium:

\(\begin{array}{*{20}{c}}{{K_p}}&{ = {{\left( {{P_{{H_2}{\rm{O}}}}} \right)}^6}}\\{{P_{{{\rm{H}}_2}{\rm{O}}}}}&{ = \sqrt[6]{{{K_p}}}}\\{}&{ = \sqrt[6]{{5.09 \times {{10}^{ - 44}}}}}\\{}&{}\\{}&{ = 6.09 \times {{10}^{ - 8}}{\rm{atm}}}\end{array}\)

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Most popular questions from this chapter

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 掳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) 鈬� A鈭�(aq) + H+(aq)]or HB[HB(aq) 鈬� H+(aq) + B鈭�(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鈭� or B鈭�), 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.)

Nitrogen and oxygen react at high temperatures.

(a) Write the expression for the equilibrium constant \(\left( {{K_c}} \right)\)for the reversible reaction

\(\Delta H = 181kJ\)

(b) What will happen to the concentrations of \({N_2},{O_2}, and\;NO\)at equilibrium if more \({O_2}\)is added?

(c) What will happen to the concentrations of \({N_2},{O_2}, and\;NO\)at equilibrium if \({N_2}\)is removed?

(d) What will happen to the concentrations of \({N_2},{O_2}, and\;NO\) at equilibrium if \(NO\)is added?

(e) What will happen to the concentrations of \({N_2},{O_2}, and\;NO\)at equilibrium if the pressure on the system is increased by reducing the volume of the reaction vessel?

(f) What will happen to the concentrations of \({N_2},{O_2}, and\;NO\) at equilibrium if the temperature of the system is increased?

(g) What will happen to the concentrations of \({N_2},{O_2}, and\;NO\) at equilibrium if a catalyst is added?

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.

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

\((a)\,{N_2}\left( g \right) + 3{H_2}\left( g \right)\rightleftharpoons 2N{H_3}(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_C} = 0.50\,\,at\,\,400^\circ C\)

\((b){{\rm{H}}_2}(g) + {{\rm{I}}_2}(g)\rightleftharpoons 2HI(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_c} = 50.2\,at\,{448^\circ }{\rm{C}}\)

\((c)N{a_2}{\rm{S}}{{\rm{O}}_4} \cdot 10{{\rm{H}}_2}O(s)\rightleftharpoons N{a_2}{\rm{S}}{{\rm{O}}_4}(s) + 10{{\rm{H}}_2}O(g){K_P} = 4.08 \times {10^{ - 25}}at\,{25^\circ }{\rm{C}}\)

\((d){{\rm{H}}_2}{\rm{O}}(l)\rightleftharpoons {{\rm{H}}_2}{\rm{O}}(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_P} = 0.122\,at\,{50^\circ }{\rm{C}}\)

Write the expression of the reaction quotient for the ionization of HOCN in water.
See all solutions

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