Chapter 14, Problem 14.81QA

Interpretation Introduction

To find:

a) Calculate the equilibrium partial pressures of the reactant and products using the given initial partial pressures of $PC{l}_5$ and $PC{l}_3$.

b) Explain the effect of addition of more $C{l}_2$ after equilibrium is reached on the concentrations of $PC{l}_5$ and $PC{l}_3$.

Answer to Problem 14.81QA

Solution:

a) Equilibrium partial pressure of $PC{l}_5=0.024 atm, PC{l}_3=1.036 atm and C{l}_2=0.536 atm.$

b) If we add more $C{l}_2$ after equilibrium is reached, the concentration of $PC{l}_3$ will decrease and the concentration of $PC{l}_5$ will increase.

Explanation of Solution

1) Concept:

We need to draw the RICE table for the given reaction and initial partial pressure. Then, we can write the $K_p$ expression and calculate the value of equilibrium partial pressures of all reactants and products.

2) Formula:

$K_p=\frac{{\left[P_{products}\right]}^{coefficents}}{{\left[P_{reactants}\right]}^{coefficents}}$

Here, $K_p$ is the equilibrium partial pressure constant, $\left[P_{products}\right]$ is the partial pressure of products and $\left[P_{reactants}\right]$ is the partial pressure of reactants and the coefficient are the coefficients of the reactants and products from the balanced reaction.

3) Given:

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

$Initial partial pressure: P_{PC{l}_5}=0.560 atm and P_{PC{l}_3}=0.500 atm$

$T=500 K$

$K_p=23.6$

4) Calculations:

a) RICE table for given reaction:

Reaction	$PC{l}_{5\left(g\right)}\rightleftharpoons PC{l}_{3\left(g\right)} + C{l}_{2\left(g\right)}$
	$P_{PC{l}_5}\left(atm\right)$	$P_{PC{l}_3}\left(atm\right)$	$P_{C{l}_2}\left(atm\right)$
Initial	$0.560$	$0.500$	$0$
Change	$-x$	$+x$	$+x$
Equilibrium	$(0.560-x)$	$(0.500+x)$	$x$

$K_p=\frac{{\left[P_{products}\right]}^{coefficents}}{{\left[P_{reactants}\right]}^{coefficents}}$

$K_p=\frac{{\left[P_{PC{l}_3}\right]}^1{\left[P_{C{l}_2}\right]}^1}{{\left[P_{PC{l}_5}\right]}^1}$

$23.6=\frac{\left(0.500+x\right)\left(x\right)}{(0.560-x)}$

$23.6\times (0.560-x)=\left(0.500+x\right)\left(x\right)$

$13.216-23.6x=0.5x+x^2$

$0.5x+x^2-13.216+23.6x=0$

$x^2+24.1x-13.216=0\dots \left(1\right)$

This equation fits the general form of a quadratic equation:

$a{x}^2+bx+C=0$

We need to solve this quadratic equation $\left(1\right)$ using the following equation:

$x=\frac{-b \pm \surd (b^2-4ac)}{2a}\dots \left(2\right)$

Here $\mathrm{a}\mathrm{ }\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{b}$ are coefficients of ${\mathrm{x}}^2,\mathrm{ }\mathrm{x}\mathrm{ }$ respectively and $c$ is a constant term in equation $\left(1\right).$

Therefore,$a=1, b=24.1 and c=-13.216$

Plugging these values in equation $\left(2\right),$ we get:

$x=\frac{-24.1 \pm \surd \left[{\left(24.1\right)}^2-(4\times 1\times ( -13.216)\right]}{2\times 1}$

$x=\frac{-24.1 \pm 25.173}{2}$

Solving this equation, we can get two values of $x$ which are:  $x=0.5364 and x= -24.6364$

We will use the positive value of $x$ for further calculation because the partial pressure of the gas can never be negative.

$So,P_{PC{l}_5}=\left(0.560-x\right)=(0.560-0.5364 )=0.024 atm$

$P_{CO} =\left(0.500+x\right)=\left(0.500+0.5364 \right)=1.036 atm$

$P_{H_2}=x=0.5364 =0.536 atm$

b) After equilibrium is reached, if more $C{l}_2$ is added to the reaction mixture, the available $PC{l}_3$ will react with it to favor the reverse reaction. Therefore, the concentration of $PC{l}_5$ will increase and the concentration of $PC{l}_3$ will decrease.

Conclusion:

Using initial partial pressure of $PC{l}_3 and PC{l}_5$ we got the equilibrium concentrations of all the reactants and products. If we add more $C{l}_2$ after equilibrium reached, the concentration of $PC{l}_3$ will decrease and the concentration of $PC{l}_5$ will increase.