Chapter 14, Problem 14.88QA

Interpretation Introduction

To find:

a) Calculate the equilibrium partial pressures of the products and reactants using the given initial partial pressures of $H_{2}O$ and $CO.$

b) Calculate the equilibrium partial pressures of the products and reactants after sufficient $H_{2}O$ and $CO$ are added to the equilibrium mixture in part $\left(a\right)$ to initially increase the partial pressure of both gases by 0.075 atm.

Answer to Problem 14.88QA

Solution:

Equilibrium partial pressures are:

a)	$P_{H_{2O}}$	$0.275 atm$
	$P_{CO}$	$5.17 atm$
	$P_{H_2}$	$0.167 atm$

b)	$P_{H_{2O}}$	$0.322 atm$
	$P_{CO}$	$5.19 atm$
	$P_{H_2}$	$0.195 atm$

Explanation of Solution

1) Concept:

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

2) Formula:

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

$K_p=K_c{\left(RT\right)}^{∆n}$

Here, $K_p$ is 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 coefficient are the coefficient of reactants and product from the balanced reaction. $K_c$ is equilibrium constant with respect to concentration, $R$ is gas constant is $0.0821 L.atm/mol.K$, T is absolute temperature and $∆n$ is the

difference of total moles of gas on product side and total moles of gas on reactant side.

3) Given:

$H_2{O}_{\left(g\right)}+C_{\left(s\right)}\leftrightarrow C{O}_{\left(g\right)}+H_{2\left(g\right)}$

i) Initial partial pressure:$P_{H_{2O}}=0.442 atm$ and $P_{CO}=5.0 atm$

ii) $T={1000}^{0}C$

iii) $K_c=3.0\times {10}^{-2}$

iv) $R= 0.0821 L.atm/mol.K$

4) Calculations:

$T={1000}^{0}C={1000}^{0}C+273.15=1273.15 K$

$∆n=\left(total moles of gas on product side\right)- \left(total moles of gas on reactant side\right)$

$∆n=\left(2-1\right)=1$

$K_p=K_c{\left(RT\right)}^{∆n}$

$K_p=3.0\times {10}^{-2}\times {\left(0.0821\frac{L.atm}{mol.K}\times 1273.15K\right)}^1$

$K_p=3.0\times {10}^{-2}\times 104.5256$

$K_p=3.13576$

a)

RICE table for the given reaction:

Reaction	$H_2{O}_{\left(g\right)} + C_{\left(s\right)}\rightleftharpoons C{O}_{\left(g\right)} + H_{2\left(g\right)}$
	$P_{H_{2O}} \left(atm\right)$		$P_{CO} \left(atm\right)$	$P_{H_2} \left(atm\right)$
Initial	$0.442$		$5.0$	$0$
Change	$-x$		$+x$	$+x$
Equilibrium	$(0.442-x)$		$(5.0+x)$	x

As $C$ is solid, and is in its pure state, and is also given as excess, we do not write the values of pure substances in the RICE table and $K$ expression.

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

$K_p=\frac{{\left[P_{CO}\right]}^1{\left[P_{H_2}\right]}^1}{{\left[P_{H_{2O}}\right]}^1}$

$3.13576=\frac{\left(5.0+x\right)\left(x\right)}{(0.442-x)}$

$3.13576\times (0.442-x)=\left(5.0+x\right)\left(x\right)$

$1.386-3.13576x=5.0x+x^2$

$5.0x+x^2-1.386+3.13576x=0$

$x^2+8.31576x-1.386=0\dots \left(1\right)$

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,$a=1, b=8.31576 and c=-1.386$

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

$x=\frac{-8.31576 \pm \surd \left[{\left(8.31576\right)}^2-(4\times 1\times ( -1.386)\right]}{2\times 1}$

$x=\frac{-8.31576 \pm \surd \left(74.6958\right)}{2}$

Solving this equation we can get two values of $x$ that are:  $x=0.166934 and x= -8.30269$

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

So,$P_{H_{2O}}=\left(0.442-x\right)=(0.442-0.166934 )=0.275 atm$

$P_{CO} =\left(5.0+x\right)=\left(5.0+0.166934 \right)=5.17 atm$

$P_{H_2}=x=0.166934=0.167 atm$

Therefore, the partial pressure of reactant and products is as follows.

$P_{H_{2O}}$	$0.275 atm$
$P_{CO}$	$5.17 atm$
$P_{H_2}$	$0.167 atm$

b)

In this part, both the initial partial pressures increase by $0.075 atm$, so the new initial partial pressures are:

$P_{CO}=5.17 atm+0.075 atm=5.2419 atm$

$P_{H_2}=0.167 atm+0.075 atm=0.2419 atm$

RICE table for the given reaction:

Reaction	$H_2{O}_{\left(g\right)} + C_{\left(s\right)} \leftrightarrow C{O}_{\left(g\right)} + H_{2\left(g\right)}$
	$P_{H_{2O}} \left(atm\right)$		$P_{CO} \left(atm\right)$	$P_{H_2} \left(atm\right)$
Initial	$0.2751$		$5.2419$	$0.2419$
Change	$+x$		$-x$	$-x$
Equilibrium	$(0.2751+x)$		$(5.2419-x)$	$0.2419-x$

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

$K_p=\frac{{\left[P_{CO}\right]}^1{\left[P_{H_2}\right]}^1}{{\left[P_{H_{2O}}\right]}^1}$

$3.13576=\frac{\left(5.2419-x\right)(0.2419-x)}{(0.2751+x)}$

$3.13576\times (0.2751+x)=\left(5.2419-x\right)(0.2419-x)$

$0.86265+3.13576x=1.268-5.4838x+x^2$

$x^2-8.6196x+0.40537=0\dots \left(1\right)$

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 $x^2, x$ respectively. And $c$ is a constant term in equation $\left(1\right).$

Therefore,$a=1, b=8.6196 and c=-1.6212$

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

$x=\frac{-(-8.6196) \pm \surd \left[{\left(-8.6196\right)}^2-(4\times 1\times ( 0.4054)\right]}{2\times 1}$

$x=\frac{8.6196 \pm \surd \left(72.6753\right)}{2}$

Solving this equation, we can get two values of $x$ that are:  $x=0.04729 and x= 8.5723$

Here both values are positive. $x= 8.5723$ is a larger value than the initial concentration, therefore, the true value of x is 0.04729.

$So, P_{H_{2O}}=\left(0.2751+x\right)=(0.2751+0.04729 )=0.3224 atm$

$P_{CO}=\left(5.2419-x\right)=\left(5.2419-0.04729\right)=5.1946atm$

$P_{H_2}=(0.02419-x)=(0.02419-0.04729) =0.1946 atm$

Therefore, the partial pressure of the reactant and products is as follows.

$P_{H_{2O}}$	$0.322 atm$
$P_{CO}$	$5.19 atm$
$P_{H_2}$	$0.195 atm$

Conclusion:

Using the RICE table and the equilibrium constant expression for partial pressures, the value of x is calculated. Using the value of x, the partial pressures of all reactant and products is calculated.