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Chapter 13: Q82E (page 760)

Question: Calculate the equilibrium concentrations that result when 0.25 M O2 and 1.0 M HCl react and come to equilibrium.

Short Answer

Expert verified

The equilibrium concentration is

Step by step solution

01

Determine change in concentration:

Given information:


  • The concentration of O2 is 0.25M
  • The concentration of HCL is 1.0M
  • The equilibrium constant is

We have to find the equilibrium concentrations.

Sinceis large, we will assume that the reaction goes to completion and then comes back to equilibrium.

We will assume that the volume of a solution is 1 L, hence the number of moles of O2 is 0.25 mol and the number of moles of HCL is 1.0 mol

Since 4 moles of HCL reacts with 1 mole of O2, to produce 2 moles of Cl2 and 2 moles of H2O 1.0 mole of HCL will react with 0.25 mole of O2, and produce 0.50 moles of Cl2 (0.50 M) and 0.50 moles of H20 (0.50 M)

We have to determine the value of x,

Since is too large, we will neglect the change in concentration of Cl2, and H2O, and get

02

Determine equilibrium concentration of all species:

The change in the concentration obtained

Therefore, the equilibrium concentrations are

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Most popular questions from this chapter

Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas.

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

\(F{e_2}{O_3}(s) + 3{H_2}(g) \rightleftharpoons 2Fe(s) + 3{H_2}O(g)\) \(\Delta H = 98.7kJ\)

(b) What will happen to the concentration of each reactant and product at equilibrium if more \(Fe\)is added?

(c) What will happen to the concentration of each reactant and product at equilibrium if \({H_2}O\) is removed?

(d) What will happen to the concentration of each reactant and product at equilibrium if \({H_2}\) is added?

(e) What will happen to the concentration of each reactant and product at equilibrium if the pressure on the system is increased by reducing the volume of the reaction vessel?

(f) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?

Calcium chloride 6−hydrate, \(CaC{l_2}.6{H_2}O\), 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 \(CaC{l_2}\)?

The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium.

(a) \(\begin{aligned}{}2{\rm{N}}{{\rm{H}}_3}(\;{\rm{g}}) \rightleftharpoons {{\rm{N}}_2}(\;{\rm{g}}) + 3{{\rm{H}}_2}(\;{\rm{g}})\;\;\;{K_e} b= 17;\\\;\;\;\left( {{\rm{N}}{{\rm{H}}_3}} \right) = 0.50\,M,\;\left( {{{\rm{N}}_2}} \right) = 0.15\,M,\;\left( {{{\rm{H}}_2}} \right) = 0.12\,M\end{aligned}\)

(b) \(\begin{aligned}{}2{\rm{N}}{{\rm{H}}_3}(\;g) \rightleftharpoons {{\rm{N}}_2}(\;g) + 3{{\rm{H}}_2}(\;g)\;\;\;{K_P} = 6.8 \times {10^4};\\\;\;\;\;{\rm{initial}}\;{\rm{pressures:}}\,{\kern 1pt} {\rm{N}}{{\rm{H}}_3} = 2.00\;{\rm{atm}},\;{\kern 1pt} {\kern 1pt} {{\rm{N}}_2} = 10.00\,{\rm{atm}},\;{{\rm{H}}_2} = 10.00\,{\rm{atm}}\end{aligned}\)

(c) \(\begin{aligned}{}2{\rm{S}}{{\rm{O}}_3}(\;g) \rightleftharpoons 2{\rm{S}}{{\rm{O}}_2}(\;g) + {{\rm{O}}_2}(\;g)\;\;\;{K_c} = 0.230;\\\;{\kern 1pt} {\kern 1pt} {\kern 1pt} \;\left( {{\rm{S}}{{\rm{O}}_3}} \right) = 2.00\,M,\;\left( {{\rm{S}}{{\rm{O}}_2}} \right) = 2.00\,M,\;\left( {{{\rm{O}}_2}} \right) = 2.00\,M\end{aligned}\)

(d) \(\begin{aligned}{}2{\rm{S}}{{\rm{O}}_3}(\;g) \rightleftharpoons 2{\rm{S}}{{\rm{O}}_2}(\;g) + {{\rm{O}}_2}(\;g)\;\;\;{K_p} = 6.5\;{\rm{atm}};\\\;\;\;\;{\rm{initial}}\;{\rm{pressures:}}\;{\rm{S}}{{\rm{O}}_2} = 1.00\;{\rm{atm}},\;{{\rm{O}}_2} = 1.130\;{\rm{atm}},\;{\rm{S}}{{\rm{O}}_3} = 0\;{\rm{atm}}\end{aligned}\)

(e) \(\begin{aligned}{}2{\rm{NO}}(g) + {\rm{C}}{{\rm{l}}_2}(g) \rightleftharpoons 2{\rm{NOCl}}(g)\;\;\;{K_P} = 2.5 \times {10^3};\\\;\;\;\;{\rm{initial}}\;{\rm{pressures:}}\;{\rm{NO}} = 1.00\;{\rm{atm}},{\rm{C}}{{\rm{l}}_2} = 1.00\;{\rm{atm}},\;{\rm{NOCl}} = 0\;{\rm{atm}}\end{aligned}\)

(f) \(\begin{aligned}{}{{\rm{N}}_2}(\;g) + {{\rm{O}}_2}(\;g) \rightleftharpoons 2{\rm{NO}}(g)\;\;\;{K_c} = 0.050;\\\;\;\;\;\;\left( {{{\rm{N}}_2}} \right) = 0.100M,\;\left( {{{\rm{O}}_2}} \right) = 0.200M,\;({\rm{NO}}) = 1.00M\end{aligned}\)

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

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

Acetic acid is a weak acid that reacts with water according to this equation:

\(C{H_3}C{O_2}H(aq) + {H_2}O(aq) \rightleftharpoons {H_3}{O^ + }(aq) + C{H_3}CO_2^ - (aq)\)

Will any of the following increase the percent of acetic acid that reacts and produces \(C{H_3}CO_2^ - \)ion?

(a) Addition of \(HCl\)

(b) Addition of \(NaOH\)

(c) Addition of \(NaC{H_3}C{O_2}\)
See all solutions

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