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.

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answered: 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

answered: 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).

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