Write the reaction quotient expression for the ionization of NH3 in water:

Short Answer

Expert verified

The reaction quotient expression for the ionization of NH3 in water:

\({Q_c} = \frac{{c\left( {NH_4^ + } \right) \cdot c\left( {O{H^ - }} \right)}}{{c\left( {N{H_3}} \right) \cdot 1}}\).

Step by step solution

01

The reaction regarding the ionization of NH3 in water:

The reaction of dissociation of\({\rm{N}}{{\rm{H}}_3}\)is:

\({\rm{N}}{{\rm{H}}_3}(aq) + {{\rm{H}}_2}{\rm{O}}(l) \to {\rm{NH}}_4^ + (aq) + O{{\rm{H}}^ - }(aq)\)

(Note, there needs to be double arrow, because this is a reverse reaction).

02

The reaction quotient expression for the ionization of NH3 in water:

Equation for reaction quotient is the same as for equilibrium constant, but we don't use equilibrium concentrations, but concentrations in any time before equilibrium.

\({Q_c} = \frac{{c\left( {NH_4^ + } \right) \cdot c\left( {O{H^ - }} \right)}}{{c\left( {N{H_3}} \right)}}\)

Water is not included into the equation because it is liquid and its activity is 1,

so, it would be like this:

\({Q_c} = \frac{{c\left( {NH_4^ + } \right) \cdot c\left( {O{H^ - }} \right)}}{{c\left( {N{H_3}} \right) \cdot 1}}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

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.)

What is the pressure of \(C{O_2}\)in a mixture at equilibrium that contains \(0.50atm\)\({H_2}\), \(2.0atm\)of \({H_2}O\), and \(1.0atm\)of \(CO\) at \(99{0^0}C\)?

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

\({K_P} = 1.6\,at\, 99{0^o}C\)

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

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

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

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

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

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?

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

For which of the reactions in Exercise 13.16 does \({K_c}\) (calculated using concentrations) equal \({K_p}\)(calculated using pressures)?

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

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free