Question

For the exothermic reaction
PCl3(g)+Cl2(g)?PCl5(g)
Kp = 0.180 at a certain temperature.
A flask is charged with 0.500 atm PCl3 , 0.500 atm Cl2, and 0.300atm PCl5 at this temperature.
What are the equilibrium partial pressures of PCl3 , Cl2, and PCl5, respectively?

Answer 1

Answer:

$p_{\text{PCl3}} = p_{\text{Cl2}} = \text{0.709 atm}; p_{\text{PCl5}} = \text{0.091 atm}$

Explanation:

1. Set up an ICE table

$\begin{array}{ccccc}\text{PCl}_{3} & + & \text{Cl}_{2} &\longrightarrow & \text{PCl}_{5}\\ 0.500 & & 0.500 & & 0.300 \\ -x & & -x & & +x \\ 0.500-x & & 0.500-x & & 0.300+x \\\end{array}$

2. Write the Kp expression

$K_{\text{p}} = \dfrac{p_{\text{PCl5}}}{p_{\text{PCl3}}p_{\text{Cl2}}}$

3. Substitute the partial pressures and solve for x

$\begin{array}{rcl}\dfrac{0.300+x}{(0.500-x)(0.500-x)} &=& 0.180\\\\0.300 + x & = & 0.180(0.500-x)(0.500-x)\\0.300+x& = & 0.180(0.250 - x + x^{2})\\0.300+x & = & 0.045 - 0.180x + 0.180x^{2}\\0&= & 0.180x^{2} -1.180x -0.255\\x = \mathbf{-0.209} & & x = 6.765\\\end{array}$

We reject the positive answer, because, even if the reaction goes completely to the right, the partial pressure of PCl₅ cannot exceed 0.8 atm.

The negative sign shows that the reaction goes in the reverse direction.

$p_{\text{PCl3}} = 0.500 - x = 0.500 + 0.209 = \textbf{0.709 atm}\\p_{\text{Cl2}} = 0.500 - x = 0.500 + 0.209 = \textbf{0.709 atm}\\p_{\text{PCl5}} = 0.300 + x = 0.300 - 0.209 = \textbf{0.091 atm}$

Check:

$\begin{array}{rcl}\dfrac{0.091}{0.709\times 0.709} &=&0.180\\\\0.181 & = & 0.180\\\end{array}$

Close enough.

Answer 2

The partial pressure of phosphorus trichloride has been 0.709 atm.

The partial pressure of chloride has been 0.709 atm.

The partial pressure of phosphorus pentachloride has been 0.091 atm.

The partial pressure has been given as the pressure exerted by each gas in the mixture. The Kp for the reaction has been the ratio of the pressure of the product to reactant at equilibrium.

The given reaction has been:

$\rm PCl_3\;+\;Cl_2\;\rightarrow\;PCl_5$

Computation of Equilibrium partial pressure

The ICE table for the equilibrium pressure has been attached.

From the ICE table, the Kp of the reaction has been given as:

$Kp=\dfrac{P_{PCl_5}}{P_{PCl_3}\;\times\;P_{Cl_2}}$

Substituting the values for x:

$0.18=\dfrac{0.3+x}{(0.5-x)(0.5-x)} \\ x=-0.209$

The partial pressure of the gases at equilibrium has been:

$\rm PCl_3=0.5-\textit x\\ PCl_3=0.5-(-0.209)\\ PCl_3=0.709$

The partial pressure of phosphorus trichloride has been 0.709 atm.

$\rm Cl_2=0.5-\textit x\\Cl_2=0.5-(-0.209)\\Cl_2=0.709$

The partial pressure of chloride has been 0.709 atm.

$\rm PCl_5=0.3+\textit x\\PCl_3=0.3+(-0.209)\\PCl_3=0.091$

The partial pressure of phosphorus pentachloride has been 0.091 atm.

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