Chemistry, 29.09.2019 09:00 charae0185

The kp for the reaction below is 1.49 × 108 at 100.0°c: co(g) + cl2(g) → cocl2(g) in an equilibrium mixture of the three gases, pco = pcl2 = 2.22 × 10-4 atm. the partial pressure of the product, phosgene (cocl2), is atm.

Answers: 1

Show answers

Answers

Answer from: missbecca1229p6glel

7.34 atm

Further explanation

Given:

The Kp for the reaction below is 1.49 · 10⁸ at 100.0°C $\boxed{ \ CO_{(g)} + Cl_2_{(g)} \rightleftharpoons COCl_2_{(g)} \ }$ (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.

$\boxed{ \ K_p = \frac{(p_{(products)})^{coefficient}}{(p_{(reactants)})^{coefficient}} \ }$ $\rightarrow \boxed{ \ K_p = \frac{p COCl_2}{(p CO)(p Cl_2)} \ }$

We set p COCl₂ as the subject to be asked.

$\boxed{ \ p COCl_2 = K_p{(p CO)(p Cl_2)} \ }$

Substitute all the data above into the equation.

$\boxed{ \ p COCl_2 = 1.49 \cdot 10^8 \times 2.22 \cdot 10^{-4} \times 2.22 \cdot 10^{-4} \ }$

Thus, the partial pressure of the product, phosgene (COCl₂), is 7.34 atm.

_ _ _ _ _ _ _ _ _ _

Notes:

For the general reactions: $\boxed{ \ aA + bB \rightleftharpoons cC + dD \ }$ ,

the equilibrium constant in terms of concentrations is given by the expression:

$\boxed{ \ K_c = \frac{[C]^c[D^d]}{[A]^a[B]^b} \ }$

Learn moreWrite the equilibrium constant for the reaction Write the equilibrium constant for the reaction of a heterogeneous balance. What is [S²⁻] at equilibrium?
The kp for the reaction below is 1.49 × 108 at 100.0°c: co(g) + cl2(g) → cocl2(g) in an equilibrium

The kp for the reaction below is 1.49 × 108 at 100.0°c: co(g) + cl2(g) → cocl2(g) in an equilibrium

Answer from: nan487

The partial pressure of phosgene $\left( {{\text{COC}}{{\text{l}}_2}} \right)$ is $\boxed{7.343316{\text{ atm}}}$ .

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 ${{\text{K}}_{\text{p}}}$ .

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 partial pressures of C and D respectively.

${{\text{P}}_{\text{A}}}$ and ${{\text{P}}_{\text{B}}}$ 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:

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

The expression for equilibrium constant for this 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)

Where,

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

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

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

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

Rearrange equation (1) to calculate ${{\text{P}}_{{\text{COC}}{{\text{l}}_{\text{2}}}}}$ .

${{\text{P}}_{{\text{COC}}{{\text{l}}_2}}} = {{\text{K}}_{\text{p}}}{{\text{P}}_{{\text{CO}}}}{{\text{P}}_{{\text{C}}{{\text{l}}_2}}}$ …… (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 (2).

$\begin{aligned}{{\text{P}}_{{\text{COC}}{{\text{l}}_2}}} &= \left( {1.49 \times {{10}^8}} \right)\left( {2.22 \times {{10}^{ - 4}}{\text{ atm}}} \right)\left( {2.22 \times {{10}^{ - 4}}{\text{ atm}}} \right) \\&= 7.343316{\text{ atm}} \\\end{aligned}$

Learn more:

Sort the solubility of gas will increase or decrease: . What is the pressure of the gas?: .

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.

Answer from: Quest

why is cellular respiration important for homeostasis?

it changes food into energy that the organism can use.

which of the following is most likely to happen if a cell is not able to divide?

the cell will not be replaced when it is damaged.

explanation:

Answer from: Quest

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and conduct electricity easily while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.