Chapter 14, Problem 14.121QA

Interpretation Introduction

To find:

a) Explain if carbon dioxide is a viable source of the fuel $CO$.

b) Explain whether the reaction is endothermic.

c) Calculate the value of $K_p$ at each temperature.

d) Explain whether the decomposition of $C{O}_2$ is an antidote for global warming.

Answer to Problem 14.121QA

Solution:

a) Carbon dioxide is not a viable source of the fuel $CO$.

b) The reaction is endothermic.

c) The value of $K_p$ at each temperature:

Temperature $\left(\mathit{K}\right)$	${\mathit{K}}_{\mathit{p}}$
$1500$	$5.5\times {10}^{-11}$
$2500$	$4.0\times {10}^{-3}$
$3000$	$0.40$

d) The decomposition of $C{O}_2$ is not an antidote for global warming.

Explanation of Solution

1) Concept

Given the percentage of decomposition of CO₂ at different temperatures and partial pressure of the CO₂ at 1 atm, we can calculate the Kp and  determine whether reaction is endothermic.

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 coefficient of reactants and product from the balanced reaction.

3) Given:

i) $2C{O}_{2\left(g\right)}\rightleftharpoons 2C{O}_{\left(g\right)}+O_{2\left(g\right)}$

ii) $Initial partial pressure of C{O}_2=1 atm$

iii) Percent decomposition of $C{O}_2$ at different temperatures:

Temperature $\left(\mathit{K}\right)$	Decomposition $\left(\mathit{\%}\right)$
$1500$	$0.048$
$2500$	$17.6$
$3000$	$54.8$

4)  Calculations:

We can say that the amount of $C{O}_2$ decomposed increases with increasing temperature. This means that $K_2>K_1$ in the equation:

$\ln \frac{K_2}{K_1}= -\frac{\Delta H^0}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Where $T_2>T_1, so \left(\frac{1}{T_2}-\frac{1}{T_1}\right)<0.$

As the temperature increases, value of $K_p$ increases. Therefore,$\ln \left(\frac{K_2}{K_1}\right)>0.$

From $\left(\frac{1}{T_2}-\frac{1}{T_1}\right)<0$, the value of $∆H$ must be positive. Therefore, the given reaction is endothermic

RICE table for given reaction:

Reaction	$2C{O}_{2\left(g\right)} \rightleftharpoons 2C{O}_{\left(g\right)} + O_{2\left(g\right)}$
	$P_{C{O}_2}\left(atm\right)$	$P_{CO}\left(atm\right)$	$P_{O_2}\left(atm\right)$
Initial	$1$	$0$	$0$
Change	$-2x$	$+2x$	$+x$
Equilibrium	$(1-2x)$	2x	$x$

Change in concentration of $C{O}_2$ is $2x$, the value of x can be calculated as,

$At T=1500 K,$

$2x=\frac{0.048}{100}=0.00048$

$x=\frac{0.00048}{2}=0.00024$

$\left(1-2x\right)=(1-0.00048)=0.99952$

Therefore, equilibrium constant expression for partial pressures for the given reaction is:

$K_p=\frac{{\left(P_{CO}\right)}^2{\left(P_{O_2}\right)}^1}{{\left(P_{C{O}_2}\right)}^2}=\frac{{\left(0.00048\right)}^2\left(0.00024\right)}{{\left(0.99952\right)}^2}=5.5\times {10}^{-11}$

$At T=2500 K,$

$2x=\frac{17.6}{100}=0.176$

$x=\frac{0.176}{2}=0.088$

$\left(1-2x\right)=\left(1-0.0176\right)=0.824$

Therefore, equilibrium constant expression for partial pressures for the given reaction is:

$K_p=\frac{{\left(P_{CO}\right)}^2{\left(P_{O_2}\right)}^1}{{\left(P_{C{O}_2}\right)}^2}=\frac{{\left(0.176\right)}^2\left(0.088\right)}{{\left(0.824\right)}^2}=4.0\times {10}^{-3}$

$At, T=3000K$

$2x=\frac{54.8}{100}=0.548$

$x=\frac{0.548}{2}=0.274$

$\left(1-2x\right)=\left(1-0.548\right)=0.452$

Therefore, equilibrium constant expression for partial pressures for the given reaction is:

$K_p=\frac{{\left(P_{CO}\right)}^2{\left(P_{O_2}\right)}^1}{{\left(P_{C{O}_2}\right)}^2}=\frac{{\left(0.548\right)}^2\left(0.274\right)}{{\left(0.452\right)}^2}=0.40$

The value of the equilibrium constant increases with increasing temperature, but it is still less than 1. Thus this reaction does not favor products even at a very high temperature, and this is not a viable source of $CO$. Therefore, it is not an antidote to decrease $C{O}_2$ as a contributor to global warming.

Final statement:

Using the given $\%$ decomposition, we calculated the $K_p$ values at different temperatures.