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Chapter 13: Q64E (page 758)

Why are there no changes specified for NH4HS? What property of Ni does change?

Short Answer

Expert verified

Pure solids (or liquids) are excluded from the equilibrium expression because their effective concentrations stay constant throughout the reaction.

Since,NH4HSis solid, there will be no change in concentration

But the property ofNH4HS(s) that does change is its mass

Step by step solution

01

Properties of sodium

  • Sodium has a strong metallic luster and in color is very analogous to silver.
  • It is soft at common temperatures that it may be formed into leaves by the pressure of the fingers.
  • Sodium compounds soon tarnish on exposure to the air, though less rapidly than potassium.
  • Sodium is instantly oxidized by water, and hydrogen gas in temporary union with a little sodium being disengaged.
02

Reason for no changes

Pure solids (or liquids) are excluded from the equilibrium expression because their effective concentrations stay constant throughout the reaction.

sinceNH4HS is solid, there will be no change in concentration.

But the property ofNH4HS (S) that does change is its mass.

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Most popular questions from this chapter

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.

Question:At 1 atm and \(2{5^o}C,N{O_2}\) with an initial concentration of \(1.00M\;is\;3.3 \times 1{0^{ - 3}}\% \)decomposed into \(NO\;and\;{O_2}\). Calculate the value of the equilibrium constant for the reaction. \(2N{O_2}(g) \rightleftharpoons 2NO(g) + {O_2}(g)\)

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 \({\bf{25}}^\circ {\bf{C}}\)?

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

Question: A reaction is represented by this equation: \({K_c} = 1 \times 1{0^3}\)

(a) Write the mathematical expression for the equilibrium constant.

Question: An equilibrium is established according to the following equation

\({K_c} = 4.6\)

What will happen in a solution that is 0.20 M in each \({\rm{Hg}}_2^{2 + },{\rm{NO}}_3^ - ,{{\rm{H}}^ + },{\rm{H}}{{\rm{g}}^{2 + }}\)and \({\rm{HN}}{{\rm{O}}_2}{\rm{\;?\;}}\)

a)\({\rm{H}}{{\rm{g}}_2}^{2 + }\) will be oxidized and \({\rm{N}}{{\rm{O}}_3}\)reduced,

b) \({\rm{H}}{{\rm{g}}_2}^{2 + }\)will be oxidized and \({\rm{N}}{{\rm{O}}_3}\) oxidized.

c)\({\rm{H}}{{\rm{g}}_2}^{2 + }\)will be oxidized and \({\rm{HN}}{{\rm{O}}_2}\)reduced.

d) \({\rm{H}}{{\rm{g}}_2}^{2 + }\)will be oxidized and \({\rm{HN}}{{\rm{O}}_2}\)oxidized.

(e) There will be no change because all reactants and products have an activity of 1.
See all solutions

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