Respuesta :
Kp can be found using the formula:
P(isobutane) divided by P(butane)
which is equal to 25.
butane ⇌ isobutane
1 mole ⇌ 1 mole
x mole ⇌ x mole
Substituting into the Kp formula:
25 = (x)/(10 - x)
P(isobutane) = x atm
P(butane) = (10 - x) atm
Given:
Butane = 5 atm
Isobutane = 0 atm
Hence,
5 = 10-x
x=5
Hope it helps.
P(isobutane) divided by P(butane)
which is equal to 25.
butane ⇌ isobutane
1 mole ⇌ 1 mole
x mole ⇌ x mole
Substituting into the Kp formula:
25 = (x)/(10 - x)
P(isobutane) = x atm
P(butane) = (10 - x) atm
Given:
Butane = 5 atm
Isobutane = 0 atm
Hence,
5 = 10-x
x=5
Hope it helps.
The equilibrium partial pressures of butane and isobutene are 0.2 atm and 4.8 atm respectively.
Further Explanation:
Chemical equilibrium is the state in which concentration of reactants and products become constant and do not change with time. This is because the rate of forward and backward direction becomes equal. The general equilibrium reaction is as follows:
[tex]{\text{A}}\left({aq}\right)+{\text{B(}}aq{\text{)}}\rightleftharpoons {\text{C(}}aq{\text{)}}+{\text{D(}}aq{\text{)}}[/tex]
Equilibrium constant is the constant that relates the concentration of product and reactant at equilibrium. The formula to calculate the equilibrium constant for general reaction is as follows:
[tex]{\text{K}}=\frac{{\left[ {\text{D}}\right]\left[{\text{C}}\right]}}{{\left[ {\text{A}} \right]\left[{\text{B}} \right]}}[/tex]
Here, K is the equilibrium constant.
For the gaseous reaction,
[tex]{\text{A(g)}}+{\text{B(g)}}\rightleftharpoons {\text{C(g)}}+{\text{D(g)}}[/tex]
When reactants and products are gases instead of liquid then the equilibrium is symbolized as [tex]{{\text{K}}_{\text{p}}}[/tex] . [tex]{{\text{K}}_{\text{p}}}[/tex] is written in terms of partial pressure of each gas as follows:
[tex]{{\text{K}}_{\text{p}}}=\frac{{{P_C}{P_D}}}{{{P_A}{P_B}}}[/tex] ......(1)
The given reaction is an isomerization of butane and isobutene,
[tex]{\text{butane }}\rightleftharpoons {\text{ isobutane}}[/tex]
The expression of [tex]{{\text{K}}_{\text{p}}}[/tex] for the above reaction is,
[tex]{{\text{K}}_{\text{p}}} = \frac{{{P_{{\text{isobutane}}}}}}{{{P_{{\text{butane}}}}}}[/tex] ......(2)
The value of[tex]{{\text{K}}_{\text{p}}}[/tex] is 25.
The initial concentration of butane is 5 atm and initial concentration of isobutene is zero.
Let the change in pressure at equilibrium be x. Therefore, the partial pressure of butane becomes 5-x at equilibrium. The partial pressure of isobutane becomes x at equilibrium.
Substitute x for [tex]{P_{{\text{isobutane}}}}[/tex], 5-x for [tex]{P_{{\text{butane}}}}[/tex] in equation (2).
[tex]{{\text{K}}_{\text{P}}} = \frac{{\text{x}}}{{5 - {\text{x}}}}[/tex] ......(3)
Rearrange the equation (3) and substitute 25 for [tex]{{\text{K}}_{\text{p}}}[/tex] to calculate value of x.
[tex]\begin{aligned}{\text{x}}&=\frac{5}{{\left({\frac{1}{{{{\text{K}}_{\text{P}}}}} + 1}\right)}}\\&=\frac{5}{{\left( {\frac{1}{{{\text{25}}}} + 1}\right)}}\\&=4.8{\text{ atm}}\\\end{aligned}[/tex]
The equilibrium partial pressure of butane is,
[tex]\begin{aligned}{P_{{\text{butane}}}}&=\left({5 - {\text{x}}}\right){\text{ atm}}\\&=\left( {5 - 4.8}\right){\text{ atm}}\\&=0.2{\text{ atm}}\\\end{aligned}[/tex]
The equilibrium partial pressure of isobutane is equal to x, therefore, [tex]{P_{{\text{isobutane}}}}[/tex] is 4.8.
Learn more:
1. Calculation of equilibrium constant of pure water at 25°c: https://brainly.com/question/3467841
2. Complete equation for the dissociation of [tex]{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}[/tex] (aq): https://brainly.com/question/5425813
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Equilibrium
Keywords: Isomerization reaction, butane, isobutene, Kp=25, the initial pressures of butane and isobutane
