Chapter 14, Problem 14.87QA

Interpretation Introduction

To calculate:

The equilibrium partial pressures of $\mathrm{C}\mathrm{O}$ and $\mathrm{C}{\mathrm{O}}_2$.

Answer to Problem 14.87QA

Solution:

Equilibrium partial pressure of $\mathrm{C}\mathrm{O}\mathrm{ }\mathrm{i}\mathrm{s}\mathrm{ }2.4\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{m}$ and that of ${\mathrm{C}\mathrm{O}}_2\mathrm{ }\mathrm{i}\mathrm{s}\mathrm{ }3.8\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{m}$.

Explanation of Solution

1) Concept:

Here we are using equilibrium constant expression of partial pressure given below using the equilibrium partial pressures of $\mathrm{C}\mathrm{O}\mathrm{ }\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{C}{\mathrm{O}}_2$ from the RICE table.

2) Formula:

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

Here, ${\mathrm{K}}_{\mathrm{p}}$ is equilibrium partial pressure constant, $\left[{\mathrm{P}}_{\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{s}}\right]$ is the partial pressure of products, $\left[{\mathrm{P}}_{\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{s}}\right]$ is the partial pressure of reactants, and coefficient are the coefficient of reactants and product from the balanced reaction.

3) Given:

i) $K_p=1.5$

ii) $T=700^{0}C$

iii) $Reaction:C{O}_{2\left(g\right)}+C_{\left(s\right)}\leftrightarrow 2C{O}_{\left(g\right)}$

iv) $Initial: P_{C{O}_2}=5.0 atm and P_{CO }=0.0 atm$

4) Calculations:

RICE table for the given reaction:

Reaction	$C{O}_{2\left(g\right)} + C_{\left(s\right)} \leftrightarrow 2C{O}_{\left(g\right)}$
	$P_{C{O}_2}\left(atm\right)$		$P_{CO }\left(atm\right)$
Initial	$5.0$		$0.0$
Change	$-x$		$+2x$
Equilibrium	$(5.0-x)$		$(0.0-2x)$

$\mathrm{C}$ is solid and in its pure state, and we do not write the values of pure substances in the RICE table and $\mathrm{K}$ expression.

$K_p=\frac{{\left[P_{CO}\right]}^2}{{\left[P_{C{O}_2}\right]}^1}$

$1.5=\frac{{\left[2x\right]}^2}{{\left[5.0-x\right]}^1}$

$1.5=\frac{4x^2}{(5.0-x)}$

$1.5\times \left(5.0-x\right)=4x^2$

$4x^2+1.5 x-7.5=0\dots \left(1\right)$

This equation fits the general form of a quadratic equation.

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

We need to solve this quadratic equation (1) 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 coefficient of ${\mathrm{x}}^2,\mathrm{ }\mathrm{x}\mathrm{ }$ respectively. And c is constant term in equation (1).

Therefore,$a=4, b=1.5 and c=-7.5$

Plugging these values in equation (2), we can get:

$x=\frac{-1.5 \pm \surd \left[{\left(1.5\right)}^2-(4\times 4\times ( -7.5)\right]}{2\times 4}$

$x=\frac{-1.5 \pm \surd \left[122.25\right]}{8}$

Solving this equation, we get two values of $\mathrm{x}\mathrm{ }$ as  $\mathrm{x}=1.19458\mathrm{ }\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{x}=\mathrm{ }-1.5694$

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

$\left[CO\right]=2x=2*1.19458=2.4 atm$

$\left[C{O}_2\right]=\left(5.0-x\right)=\left(5.0-1.19458\right)=3.8 atm$

Therefore, equilibrium partial pressures of $\left[CO\right]$ and $\left[C{O}_2\right]$ are $2.4 atm$ and $3.8 atm$ respectively.

Conclusion:

Equilibrium partial pressures of the gaseous reactant and product can be determined using the gas phase equilibrium constant expression (K_P) and the ICE table.