Question

The equilibrium constant (kp) for the interconversion of pcl5 and pcl3 is 0.0121: pcl5 (g) ⇌ pcl3 (g) cl2 (g) a vessel is charged with pcl5 giving an initial pressure of 0.123 atm. at equilibrium, the partial pressure of pcl3 is ________ atm.

Answer 1

The partial pressure of ${\text{PC}}{{\text{l}}_3}$ is $\boxed{{\text{0}}{\text{.033 atm}}}$.

Further explanation:

Chemical equilibrium is established whenever the rate of forward reaction and that of backward reaction becomes equal. At this condition, concentration of both reactants and products become constant.

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

Given reaction is as follows:

${\text{PC}}{{\text{l}}_{\text{5}}}\left( {\text{g}} \right) \rightleftharpoons {\text{PC}}{{\text{l}}_{\text{3}}}\left( {\text{g}} \right) + {\text{C}}{{\text{l}}_{\text{2}}}\left( {\text{g}} \right)$  

The equilibrium constant for this reaction is calculated by following formula:

${{\text{K}}_{\text{p}}} = \dfrac{{{{\text{P}}_{{\text{PC}}{{\text{l}}_3}}}{{\text{P}}_{{\text{C}}{{\text{l}}_2}}}}}{{{{\text{P}}_{{\text{PC}}{{\text{l}}_5}}}}}$                                                      ...... (1)

Here,

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

${{\text{P}}_{{\text{PC}}{{\text{l}}_3}}}$ is partial pressure of ${\text{PC}}{{\text{l}}_3}$ .

${{\text{P}}_{{\text{PC}}{{\text{l}}_5}}}$ is partial pressure of ${\text{PC}}{{\text{l}}_5}$.

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

Consider x to be change in equilibrium pressure. Therefore, equilibrium pressure of ${\text{PC}}{{\text{l}}_3}$ and ${\text{C}}{{\text{l}}_2}$ after decomposition of ${\text{PC}}{{\text{l}}_{\text{5}}}$ becomes x and x respectively and that of ${\text{PC}}{{\text{l}}_{\text{5}}}$ becomes (0.123 – x).

Substitute 0.0121 for ${{\text{K}}_{\text{p}}}$ , x for ${{\text{P}}_{{\text{C}}{{\text{l}}_{\text{2}}}}}$ and x for ${{\text{P}}_{{\text{PC}}{{\text{l}}_3}}}$ and (0.123 – x) for ${{\text{P}}_{{\text{PC}}{{\text{l}}_5}}}$ in equation (1).

${\text{0}}{\text{.0121}} = \dfrac{{{x^2}}}{{\left( {0.123 - x} \right)}}$  

Solving for x,

$x = - 0.0451$  

Or,

$x = 0.033$

 

Since pressure can never be negative, x = -0.0451 is rejected. Therefore the value of x comes out to be 0.33.

Hence partial pressure of ${\text{PC}}{{\text{l}}_{\text{3}}}$ is 0.033 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: Senior School

Subject: Chemistry

Chapter: Chemical equilibrium

Keywords: Kp, chemical equilibrium, PCl3, Cl2, PCl5, 0.033 atm, partial pressure, x, 0.0121, 0.123 – x.

Answer 2

Let's assume that change of pressure of PCl₅ is X

According to the ICE table,
                               PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)
Initial Pressure       0.123           -             -
Change                   -X               +X          +X
Equilibrium             0.123 - X       X           X

       Kp = Ppcl₃(g) x Pcl₂(g) / Ppcl₅(g)
by substitution,
0.0121 = X × X / (0.123 - X)
0.0121 × (0.123 - X) = X²
1.4883 × 10⁻³ - 0.0121X = X²
X² + 0.0121X - 1.4883 × 10⁻³  = 0
X₁ = 0.03735 or X₂ = -0.03865
X cannot be a negative value since X is a pressure.
Hence,
   X = 0.03735 atm

Hence, partial pressure of PCl₃ is 0.03735 atm