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Question: Butane exists as two isomers, n−butane and isobutane.

\({K_P} = 2.5\;at\;2{5^o}C\)

What is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of 1.22 atm?

Short Answer

Expert verified

The pressure of isobutane is 0.87 atm.

Step by step solution

01

Given Data

Given information:

\({\text{n - butane}} \rightleftharpoons {\text{isobutane}}\)

  • The total pressure at equilibrium is 1.22 atm
  • The equilibrium constant is Kp=2.5

We have to find the pressure of isobutane (x) in a container.

\(\begin{array}{*{20}{c}}{{P_{n - {\rm{\;butane\;}}}} + {P_{{\rm{isobutane\;}}}} = 1.22{\rm{atm}}}\\{{P_{{\rm{isobutane\;}}}} = x}\\{{P_{n - {\rm{\;butane\;}}}} = 1.22{\rm{amt}} - x}\end{array}\)

02

Determine the value of x:

Now, we will find the value of x

\begin{array}{*{20}{c}}{{K_p} = \frac{{{P_{{\rm{isobutane\;}}}}}}{{{P_{n - {\rm{\;butane\;}}}}}}}\\{2.5 = \frac{x}{{1.22 - x}}}\\{x = 3.05 - 2.5x}\\{3.5x = 3.05}\\{x = 0.87{\rm{atm}}}\end{array}

Therefore, the pressure of isobutane is

\({P_{{\rm{isobutane\;}}}} = x = 0.87{\rm{\;atm\;}}\)

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Most popular questions from this chapter

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 \(\left[ {C{O_2}} \right] = 0.100M\)

Question: \(\;A\;0.72 - mol\)sample of \(PC{l_5}\)is put into a \(1.00 - L\) vessel and heated. At equilibrium, the vessel contains \(0.40mol\) of \(PC{l_3}(g)\) and \(0.40mol\;of\;C{l_2}(g)\). Calculate the value of the equilibrium constant for the decomposition of \(PC{l_5}\;to\;\)\(PC{l_3}\;and\;C{l_2}\)at this temperature.

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 \(\left[ {{O_2}} \right] = 0.350M\)?

Question : A 0.010Msolution of the weak acid HA has an osmotic pressure (see chapter on solutions and colloids) of 0.293 atm at 25 °C. A 0.010Msolution of the weak acid HB has an osmotic pressure of 0.345 atm under the same conditions.

(a) Which acid has the larger equilibrium constant for ionization

HA[HA(aq) ⇌ A−(aq) + H+(aq)]or HB[HB(aq) ⇌ H+(aq) + B−(aq)]?

(b) What are the equilibrium constants for the ionization of these acids?

(Hint: Remember that each solution contains three dissolved species: the weak acid (HA or HB), the conjugate base (A− or B−), and the hydrogen ion (H+). Remember that osmotic pressure (like all colligative properties) is related to the total number of solute particles. Specifically for osmotic pressure, those concentrations are described by molarities.)

Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium?

\({\rm{S}}{{\rm{O}}_2}{\rm{C}}{{\rm{l}}_2}(g)\rightleftharpoons {\rm{S}}{{\rm{O}}_2}(g) + {\rm{C}}{{\rm{l}}_2}(g)\)

\(\left( {{\rm{S}}{{\rm{O}}_2}{\rm{C}}{{\rm{l}}_2}} \right) = 0.12\;{\rm{M}},\;\left( {{\rm{C}}{{\rm{l}}_2}} \right) = 0.16\;{\rm{M and }}\left( {{\rm{S}}{{\rm{O}}_2}} \right) = 0.050\;{\rm{M}}.\;{K_c}\) for the reaction is 0.078.

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