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Chapter 13: Q69E (page 759)

Cobalt metal can be prepared by reducing cobalt (II) oxide with carbon monoxide.

\(CoO(s) + CO(g) \rightleftharpoons Co(s) + C{O_2}(g)\)

\({K_c} = 4.90 \times 1{0^2}at55{0^o}C\)

What concentration of \(CO\) remains in an equilibrium mixture with \(\left[ {C{O_2}} \right] = 0.100M\)

Short Answer

Expert verified

The concentration of CO at equilibrium is \(2.04 \times 1{0^{ - 4}}M\)

Step by step solution

01

Given information

\(CoO(s) + CO(g) \rightleftharpoons Co(s) + C{O_2}(g)\)

  1. Value of equilibrium constant \({K_C} = 4.90 \times 1{0^2}\)
  2. The concentration of \(C{O_2}\) is \(0.100\,M\)

The value of equilibrium molar concentration needs to be calculated using the equation relating equilibrium constant and molar concentrations.

02

Calculate concentration of CO at equilibrium

\(\begin{array}{*{20}{c}}{{K_c}}&{ = \frac{{\left[ {{\rm{C}}{{\rm{O}}_2}} \right]}}{{[{\rm{CO}}]}}}\\{[CO]}&{ = \frac{{\left[ {C{O_2}} \right]}}{{{K_c}}}}\\{}&{ = \frac{{0.100}}{{4.90 \times {{10}^2}}}}\\{}&{ = 2.04 \times {{10}^{ - 4}}{\rm{M}}}\end{array}\)

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Most popular questions from this chapter

How can the pressure of water vapor are increased in the following equilibrium?

\({H_2}O(l) \rightleftharpoons {H_2}O(g)\) \(\Delta H = 41kJ\)

A student solved the following problem and found the equilibrium concentrations to be \(\left[ {S{O_2}} \right] = 0.590M\), \(\left[ {{O_2}} \right] = 0.0450M\), and \(\left[ {S{O_3}} \right] = 0.260M\). How could this student check the work without reworking the problem? The problem was: For the following reaction at \(60{0^0}C\):

\(2S{O_2}(g) + {O_2}(g) \rightleftharpoons 2S{O_3}(g)\)

\({K_c} = 4.32\)

What are the equilibrium concentrations of all species in a mixture that was prepared with \(\left[ {S{O_3}} \right] = 0.500M\), \(\left[ {S{O_2}} \right] = 0M\)and \(\left[ {{O_2}} \right] = 0.350M\)?

Write the mathematical expression for the reaction quotient, \({Q_c}\), for each of the following reactions:

(a) \({N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g)\)

(b) \(4N{H_3}(g) + 5{O_2}(g) \rightleftharpoons 4NO(g) + 6{H_2}O(g)\)

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

(d) \(C{O_2}(g) + {H_2}(g) \rightleftharpoons CO(g) + {H_2}O(g)\)

(e) \(N{H_4}Cl(s) \rightleftharpoons N{H_3}(g) + HCl(g)\)

(f) \(2\;Pb{\left( {N{O_3}} \right)_2}(s) \rightleftharpoons 2PbO(s) + 4N{O_2}(g) + {O_2}(g)\)

(g) \(2{H_2}(g) + {O_2}(g) \rightleftharpoons 2{H_2}O(l)\)

(h) \({S_8}(g) \rightleftharpoons 8\;S(g)\)

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.

For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be 100%.

Is \({K_c} > 1,\; < 1\), or \( \approx 1\) for a titration reaction?
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