Chapter 10, Problem 10.110QA

Interpretation Introduction

To find:

The partial pressure of $CO$

Answer to Problem 10.110QA

Solution:

The partial pressure of $CO$ is $0.320 atm$

Explanation of Solution

1) Concept:

We are asked to find the partial pressure of $CO$ gas in the reaction mixture. We need to find first the limiting reactant. Then, we need to find moles of gaseous products $CO$ and $C{O}_2$ that are formed during the reaction and moles of unreacted ${ O}_2$. Combining the moles will give us the total moles of gaseous species. Using the total moles of gaseous products and moles of $O$, we can find mole fraction of $CO$.

According to Dalton’s law of partial pressure, the partial pressure of an individual is equal to the product of its mole fraction and total pressure.

i.e.,

$P_i={\chi }_i\times P_{total}$

where, $P_i$= partial pressure of component $i$, ${\chi }_i$= mole fraction of component $i$

$P_{total}$= total pressure.

Mole fraction is the ratio of moles of individual component to the total moles in the mixture.

i.e.

${\chi }_A=\frac{mol of A}{(mol of A+mol of B)}=\frac{mol of A}{total mol}$

2) Formula:

i) $P_i={\chi }_i\times P_{total}$

ii) ${\chi }_A=\frac{mol of A}{(mol of A+mol of B)}$

3) Given:

i) $0.156 mol C$

ii) $0.117 mol O_2$

iii) $V=10.0 L$

iv) $T=500 K$

v) $P_{total}=0.640 atm$

4) Calculation:

i) The balanced equation for the reaction is

$4C + 3O_2 \to 2C{O}_2 + 2 CO$

Finding the limiting reactant as follows:

$0.156 mol C \times \frac{3 mol O_2}{4 mol C}=0.117 mol$

As the two reactants are in stoichiometric amounts, both are completely consumed. So, after the reaction is complete, only gaseous products are present, which will govern the total pressure.

ii) From the stoichiometry of the reaction, one mole of $C$ produces two moles of $C{O}_2$, so converting moles of $C$ into moles of $C{O}_2$ as

$0.156 mol C\times \frac{1 mol C{O}_2 }{2 mol C}=0.0780 mol C{O}_2$

From the stoichiometry of the reaction, one mole of $C$ produces two moles of $C{O}_{ }$, so converting moles of $C$ into moles of $C{O}_{ }$ as

$0.156 mol C\times \frac{1 mol CO }{2 mol C}=0.0780 mol CO$

Calculating the mole fraction of $CO$ as

${\chi }_{CO}=\frac{mol of CO}{\left(mol of CO+mol of C{O}_2\right)}$

${\chi }_{CO}=\frac{0.0780\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }}{(0.0780\mathrm{m}\mathrm{o}\mathrm{l}+0.0780\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ })}=0.500$

Calculating the partial pressure of $CO$

$P_{CO}={\chi }_{COi}\times P_{total}$

$P_{CO}=0.500\times 0.640 atm=0.320 atm$

Conclusion:

Dalton’s law of partial pressure is used to find partial pressure of each of the gaseous compound in a given reaction.