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Question: A 1.00-L vessel at 400 °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?

Short Answer

Expert verified

Since initial concentration of hydrogen is \(0.50{\rm{\;mol,\;}}\)and we have to remove \(0.634{\rm{mol}}\) the equilibrium concentration cannot be \(1.1{\rm{M}}\)

Step by step solution

01

Definition

Chemical equilibrium is the state in which concentration of both thereactants andproducts have no further tendency to change with time.

Nitrogen is thechemical element with thesymbol N andatomic number 7. Nitrogen is anonmetal and the lightest member ofgroup 15 of the periodic table, often called the pnictogens.

Hydrogen is thechemical element with thesymbol H andatomic number 1. Hydrogen is the lightest element. Atstandard conditions hydrogen is agas ofdiatomic molecules having theformula H2.

02

Calculate the equilibrium constant

Given information

The volume of the vessel is 1.00L

At equilibrium constants

\(\begin{array}{*{20}{c}}{\left( {{{\rm{N}}_2}} \right) = 1.00{\rm{M}}}\\{\left( {{{\rm{H}}_2}} \right) = 0.50{\rm{M}}}\\{\left( {{\rm{N}}{{\rm{H}}_3}} \right) = 0.25{\rm{M}}}\end{array}\)

The Equillibrium constant is

\({K_c} = \frac{{{{\left| {{\rm{N}}{{\rm{H}}_3}} \right|}^2}}}{{\left| {{N_2}} \right| \times {{\left| {{H_2}} \right|}^3}}}\)

\(\frac{{{{(0.25)}^2}}}{{1.00 \times {{(0.5)}^3}}} = 0.5\)

We have to calculate the number of moles of hydrogen to be removed from the vessel to increase the concentration of nitrogen to 1.1M

03

The New equilibrium constants

The new equilibrium concentration of the\(\left| {{{\rm{N}}_2}} \right|{\rm{\;is\;}}1.1{\rm{M}}\)

And new initial concentrations are

\(\begin{array}{*{20}{c}}{ - \left( {{{\rm{N}}_2}} \right) = 1.00{\rm{M}}}\\{ - \left( {{{\rm{H}}_2}} \right) = 0.50{\rm{M}} - {\rm{x}}}\\{ - \left( {{\rm{N}}{{\rm{H}}_3}} \right) = 0.25{\rm{M}}}\end{array}\)

\({{\rm{N}}_2}({\rm{g}}) + 3{{\rm{H}}_2}({\rm{g}}) \to 2{\rm{N}}{{\rm{H}}_3}({\rm{g}})\)

Initial(M)

\(1.00\)

\(0.50 - x\)

\(0.25\)

Change(M)

\( + 0.1\)

\( + 0.3\)

\( - 0.2\)

Equillibrium(M)

\(1.1\)

\(0.80 - x\)

\(0.05\)

04

Calculate the value of X

\(\begin{array}{*{20}{c}}{{K_c} = \frac{{{{\left( {N{H_3}} \right)}^2}}}{{\left( {{N_2}} \right) \times {{\left( {{H_2}} \right)}^3}}}}\\{0.5 = \frac{{{{(0.05)}^2}}}{{1.1 \times {{(0.80 - x)}^3}}}}\end{array}\)

\(\begin{array}{*{20}{c}}{{{(0.80 - x)}^3} = 0.0045}\\{0.80 - x = 0.16565}\\{x = 0.634{\rm{M}}}\end{array}\)

Since initial concentration of hydrogen is \(0.50{\rm{\;mol,\;}}\)and we have to remove \(0.634{\rm{mol}}\) the equilibrium concentration cannot be \(1.1{\rm{M}}\)

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Most popular questions from this chapter

A sample of ammonium chloride was heated in a closed container. NH4 Cl (s)⇌ NH3 (g) + HCl(g)at equilibrium, the pressure of NH3 (g)was found to be 1.75 atm. What is the value of the equilibrium constant, Kp, for the decomposition at this temperature?

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

Round the following to the indicated number of significant figures:

(a) 0.424 (to two significant figures)

(b) 0.0038661 (to three significant figures)

(c) 421.25 (to four significant figures)

(d) 28,683.5 (to five significant figures)

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

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

For which of the reactions in Exercise 13.15 does\({K_c}\)(calculated using concentrations) equal\({K_p}\)(calculated using pressures)?

(a) \(C{H_4}(g) + C{l_2} \rightleftharpoons C{H_3}CI(g) + HCI(g)\)

(b) \({N_2}(g) + {O_2}(g)\rightleftharpoons 2NO(g)\)

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

(d) \(BaS{O_3}(s)\rightleftharpoons BaO(s) + S{O_2}(g)\)

(e) \({P_4}(g) + 5{O_2}(g)\rightleftharpoons{P_4}{O_{10}}(s)\)

(f) \(B{r_2}(\;g)\rightleftharpoons 2Br(g)\)

(g) \(C{H_4}(g) + 2{O_2}(g)\rightleftharpoons C{O_2}(g) + 2{H_2}O(l)\)

(h)\(CuS{O_4} \times 5{H_2}O(s)\rightleftharpoons CuS{O_4}(s) + 5{H_2}O(g)\)

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