Answer :
7.34 atm
Further explanation
Given:
The Kp for the reaction below is 1.49 · 10⁸ at 100.0°C[tex]\boxed{ \ CO_{(g)} + Cl_2_{(g)} \rightleftharpoons COCl_2_{(g)} \ }[/tex] (a balanced reaction)
In an equilibrium mixture of the three gases.
p CO = p Cl₂ = 2.22 · 10⁻⁴ atm.
Question:
The partial pressure of the product, phosgene (COCl₂), is __ atm.
The Process:
Let us write the equilibrium constant in terms of pressure based on the reaction above.
[tex]\boxed{ \ K_p = \frac{(p_{(products)})^{coefficient}}{(p_{(reactants)})^{coefficient}} \ }[/tex][tex]\rightarrow \boxed{ \ K_p = \frac{p COCl_2}{(p CO)(p Cl_2)} \ }[/tex]
We set p COCl₂ as the subject to be asked.
[tex]\boxed{ \ p COCl_2 = K_p{(p CO)(p Cl_2)} \ }[/tex]
Substitute all the data above into the equation.
[tex]\boxed{ \ p COCl_2 = 1.49 \cdot 10^8 \times 2.22 \cdot 10^{-4} \times 2.22 \cdot 10^{-4} \ }[/tex]
Thus, the partial pressure of the product, phosgene (COCl₂), is 7.34 atm.
_ _ _ _ _ _ _ _ _ _
Notes:
For the general reactions: [tex]\boxed{ \ aA + bB \rightleftharpoons cC + dD \ }[/tex],
the equilibrium constant in terms of concentrations is given by the expression:
[tex]\boxed{ \ K_c = \frac{[C]^c[D^d]}{[A]^a[B]^b} \ }[/tex]
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The partial pressure of phosgene [tex]\left( {{\text{COC}}{{\text{l}}_2}} \right)[/tex] is [tex]\boxed{7.343316{\text{ atm}}}[/tex] .
Further explanation:
Chemical equilibrium:
The state where reactant concentration and product concentration become constant and do not change with time is known as equilibrium.The rate of forward and backward reactions becomes equal at equilibrium.
Equilibrium constant in pressure terms:
The ratio of partial pressures of products to partial pressures of reactants, both of these terms are raised to some power equal to their respective coefficients in a balanced chemical equation. It is denoted by [tex]{{\text{K}}_{\text{p}}}[/tex].
Consider a general balanced reaction,
[tex]{\text{aA}}\left( g \right) + {\text{bB}}\left( g \right) \rightleftharpoons {\text{cC}}\left( g \right) + {\text{dD}}\left( g \right)[/tex]
The formula to calculate [tex]{{\text{K}}_{\text{p}}}[/tex] is as follows:
[tex]{{\text{K}}_{\text{p}}} = \dfrac{{{{\left[ {{{\text{P}}_{\text{C}}}} \right]}^{\text{c}}}{{\left[ {{{\text{P}}_{\text{D}}}} \right]}^{\text{d}}}}}{{{{\left[ {{{\text{P}}_{\text{A}}}} \right]}^{\text{a}}}{{\left[ {{{\text{P}}_{\text{B}}}} \right]}^{\text{b}}}}}[/tex]
Here,
[tex]{{\text{P}}_{\text{C}}}[/tex] and [tex]{{\text{P}}_{\text{D}}}[/tex] are partial pressures of C and D respectively.
[tex]{{\text{P}}_{\text{A}}}[/tex] and [tex]{{\text{P}}_{\text{B}}}[/tex] are partial pressures of A and B respectively.
a and b are stoichiometric coefficients of A and B respectively.
c and d are stoichiometric coefficients of C and D respectively.
Given reaction is as follows:
[tex]{\text{CO}}\left( {\text{g}} \right) + {\text{C}}{{\text{l}}_2}\left( {\text{g}} \right) \rightleftharpoons {\text{COC}}{{\text{l}}_{\text{2}}}\left( {\text{g}} \right)[/tex]
The expression for equilibrium constant for this reaction is as follows:
[tex]{{\text{K}}_{\text{p}}} = \dfrac{{{{\text{P}}_{{\text{COC}}{{\text{l}}_2}}}}}{{{{\text{P}}_{{\text{CO}}}}{{\text{P}}_{{\text{C}}{{\text{l}}_2}}}}}[/tex] ...... (1)
Where,
[tex]{{\text{K}}_{\text{p}}}[/tex] is equilibrium constant.
[tex]{{\text{P}}_{{\text{COC}}{{\text{l}}_{\text{2}}}}}[/tex] is partial pressure of [tex]{\text{COC}}{{\text{l}}_{\text{2}}}[/tex].
[tex]{{\text{P}}_{{\text{CO}}}}[/tex] is partial pressure of CO.
[tex]{{\text{P}}_{{\text{C}}{{\text{l}}_{\text{2}}}}}[/tex] is partial pressure of [tex]{\text{C}}{{\text{l}}_2}[/tex].
Rearrange equation (1) to calculate [tex]{{\text{P}}_{{\text{COC}}{{\text{l}}_{\text{2}}}}}[/tex].
[tex]{{\text{P}}_{{\text{COC}}{{\text{l}}_2}}} = {{\text{K}}_{\text{p}}}{{\text{P}}_{{\text{CO}}}}{{\text{P}}_{{\text{C}}{{\text{l}}_2}}}[/tex] …… (2)
Substitute [tex]1.49 \times {10^8}[/tex] for [tex]{{\text{K}}_{\text{p}}}[/tex] , [tex]2.22 \times {10^{ - 4}}{\text{ atm}}[/tex] for [tex]{{\text{P}}_{{\text{C}}{{\text{l}}_{\text{2}}}}}[/tex] and [tex]2.22 \times {10^{ - 4}}{\text{ atm}}[/tex] for [tex]{{\text{P}}_{{\text{CO}}}}[/tex] in equation (2).
[tex]\begin{aligned}{{\text{P}}_{{\text{COC}}{{\text{l}}_2}}} &= \left( {1.49 \times {{10}^8}} \right)\left( {2.22 \times {{10}^{ - 4}}{\text{ atm}}} \right)\left( {2.22 \times {{10}^{ - 4}}{\text{ atm}}} \right) \\&= 7.343316{\text{ atm}} \\\end{aligned}[/tex]
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Chemical equilibrium
Keywords: Kp, CO, Cl2, COCl2, 7.343316 atm, chemical equilibrium, pressure, partial pressure, phosgene, PCl2, PCO, PCOCl2.