Question

The Kp for the reaction below is 1.49 × 108 at 100.0°C:

Answer 1

In an equilibrium mixture of the three gases, PCO = PCl2 = 2.22 × 10-4 atm. The partial pressure of the product, phosgene (COCl2), is kp=(COCl2)/(CO)(Cl2) which is $1.48*10^=x/(2.24*10-4)^2$. So, the correct answer is A) 7.34.

Answer 2

The partial pressure of ${\text{COC}}{{\text{l}}_2}$ is $\boxed{7.34{\text{ atm}}}$.

Further explanation:

Chemical equilibrium is the state in which the concentration of reactants and products become constant and do not change with time. At this point, rate of forward and backward direction becomes equal.

Equilibrium constant in terms of pressure:

It is the ratio of partial pressures of products to that 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 ${{\text{K}}_{\text{p}}}$.

Consider a general balanced reaction,

${\text{aA}}\left( g \right) + {\text{bB}}\left( g \right) \rightleftharpoons {\text{cC}}\left( g \right) + {\text{dD}}\left( g \right)$  

The formula to calculate ${{\text{K}}_{\text{p}}}$ is as follows:

${{\text{K}}_{\text{p}}} = \dfrac{{{{\left[ {{{\text{P}}_{\text{C}}}} \right]}^{\text{c}}}{{\left[ {{{\text{P}}_{\text{D}}}} \right]}^{\text{d}}}}}{{{{\left[ {{{\text{P}}_{\text{A}}}} \right]}^{\text{a}}}{{\left[ {{{\text{P}}_{\text{B}}}} \right]}^{\text{b}}}}}$  

Here,

${{\text{P}}_{\text{C}}}$ and ${{\text{P}}_{\text{D}}}$ are the respective partial pressures of C and D.

${{\text{P}}_{\text{A}}}$ and ${{\text{P}}_{\text{B}}}$ are the respective partial pressures of A and B.

a and b are the respective stoichiometric coefficients of A and B.

c and d are the respective stoichiometric coefficients of C and D.

Given reaction occurs as follows:

${\text{CO}}\left( {\text{g}} \right) + {\text{C}}{{\text{l}}_2}\left( {\text{g}} \right) \rightleftharpoons {\text{COC}}{{\text{l}}_{\text{2}}}\left( {\text{g}} \right)$  

The formula to calculate the equilibrium constant for above reaction is as follows:

${{\text{K}}_{\text{p}}} = \dfrac{{{{\text{P}}_{{\text{COC}}{{\text{l}}_2}}}}}{{{{\text{P}}_{{\text{CO}}}}{{\text{P}}_{{\text{C}}{{\text{l}}_2}}}}}$                                                      ...... (1)

Here,

${{\text{K}}_{\text{p}}}$ is the equilibrium constant.

${{\text{P}}_{{\text{COC}}{{\text{l}}_{\text{2}}}}}$ is the pressure of ${\text{COC}}{{\text{l}}_{\text{2}}}$.

${{\text{P}}_{{\text{CO}}}}$ is the pressure of CO.

${{\text{P}}_{{\text{C}}{{\text{l}}_{\text{2}}}}}$ is the pressure of ${\text{C}}{{\text{l}}_2}$.

Substitute $1.49 \times {10^8}$ for ${{\text{K}}_{\text{p}}}$, $2.22 \times {10^{ - 4}}{\text{ atm}}$ for ${{\text{P}}_{{\text{C}}{{\text{l}}_{\text{2}}}}}$ and $2.22 \times {10^{ - 4}}{\text{ atm}}$ for ${{\text{P}}_{{\text{CO}}}}$ in equation (1).

${\text{1}}{\text{.49}} \times {\text{1}}{{\text{0}}^8} = \dfrac{{{{\text{P}}_{{\text{COC}}{{\text{l}}_2}}}}}{{\left( {2.22 \times {{10}^{ - 4}}{\text{ atm}}} \right)\left( {2.22 \times {{10}^{ - 4}}{\text{ atm}}} \right)}}$                                                       ...... (2)

Solving for ${{\text{P}}_{{\text{COC}}{{\text{l}}_{\text{2}}}}}$,

 ${{\text{P}}_{{\text{COC}}{{\text{l}}_{\text{2}}}}} = 7.34{\text{ atm}}$

Learn more:

1. Sort the solubility of gas will increase or decrease: brainly.com/question/2802008.
2. What is the pressure of the gas?:brainly.com/question/6340739.

Answer details:

Grade: School School

Subject: Chemistry

Chapter: Chemical equilibrium

Keywords: Kp, CO, Cl2, COCl2, 7.34 atm, chemical equilibrium, pressure, partial pressure.