Chapter 13, Problem 98QAP

Dinitrogen monoxide, ${\text{N}}_{\text{2}}\text{O}$, reacts with propane, ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$, to farm nitrogen, ${\text{N}}_{\text{2}}$; carbon dioxide. ${\text{CO}}_{\text{2}}$; and water, ${\text{H}}_{\text{2}}\text{O}$.

Write a balanced chemical equation for this reaction, treating all substances as gases. Include phases in your equation.

Two reservoirs, separated by a closed valve (of negligible volume), are filled with the reactants (one on each side). The pressure of ${\text{N}}_{\text{2}}\text{O}$is measured to be $\text{1}.\text{4}$atm. The pressure of ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$is measured to be $\text{1}.\text{2}$atm. Each reservoir has a capacity of $\text{1}.0\text{ L}$, and the temperature of the system is

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e valve between the reservoirs is opened, and the chemical reaction proceeds. Assuming ideal gas behavior and constant temperature, what is the partial pressure of ${\text{CO}}_{\text{2}}$after the reaction is complete?

What is the partial pressure of ${\text{N}}_{\text{2}}\text{O}$after the reaction is complete?

Interpretation Introduction

(a)

Interpretation:

To write the balanced chemical equation between nitrogen monoxide and propane that results in the production of N₂, CO₂ and H₂ O gases.

Concept Introduction:

According to Dalton’s atomic theory matter cannot be destroyed or created; in order to be consistent with this, one must make sure that the number of atoms for each distinct type of an element must be equal on both sides of a chemical reaction.

Answer to Problem 98QAP

The balanced chemical equation between nitrogen monoxide and propane that results in the production of N₂, CO₂ and H₂ O gases is:

${\text{10 N}}_2\text{O}(g){\text{ + C}}_{\text{3}}{\text{H}}_8(g)\to 10{\text{ N}}_2(g){\text{ + 3 CO}}_2(g)+{\text{ 4H}}_2\text{O}(g)$.

Explanation of Solution

Nitrogen monoxide and propane are the two reactants while N₂, CO₂ and H₂ O are the products. The chemical equation with their phases is:

${\text{ N}}_{\text{2}}\text{O}(g){\text{ + C}}_{\text{3}}{\text{H}}_8(g)\to {\text{N}}_2(g){\text{ + CO}}_2(g)+{\text{ H}}_2\text{O}(g)$

The above reaction is not balanced as the number of atoms of oxygen, carbon, and hydrogen is not same on both the sides of the reaction.

So, in order to balance the reaction, the ${\text{ N}}_{\text{2}}\text{O}$ on the reactant side is multiplied by 10, and ${\text{N}}_{\text{2}}{\text{, CO}}_{\text{2}}{\text{and H}}_{\text{2}}\text{O}$ are multiplied by 10, 3, and 4 respectively on the product side.

Thus, the balanced reaction is:

${\text{10 N}}_2\text{O}(g){\text{ + C}}_{\text{3}}{\text{H}}_8(g)\to 10{\text{ N}}_2(g){\text{ + 3 CO}}_2(g)+{\text{ 4H}}_2\text{O}(g)$.

Interpretation Introduction

(b)

Interpretation:

The partial pressure of CO₂ after the reaction undergoes completion should be determined.

Concept Introduction:

If there is a mixture of gasses in a container the total pressure exerted is the sum of partial pressure of the gasses present in the container. Partial pressure is defined as the pressure exerted by the gas if the gas was alone in the container.

$\begin{array}{l}{\text{P}}_{\text{t}\text{o}\text{t}\text{a}\text{l}}={\text{ P}}_1{\text{ + P}}_2{\text{ + P}}_3\\ \text{where }{\text{P}}_{\text{t}\text{o}\text{t}\text{a}\text{l}}=\text{total pressure}\\ {\text{P}}_{\text{i}}{\text{ =partal pressure of i}}^{\text{t}\text{h}}\text{ gas component where i=1, 2, 3}\end{array}$

Partial pressure can be derived as follows

$\begin{array}{l}{\text{P}}_{\text{A}}={\text{P}}_{\text{t}\text{o}\text{t}\text{a}\text{l}}\times {\text{Χ}}_{\text{A}}\\ {\text{where P}}_{\text{A}}=\text{partial pressure of }\text{A}\\ {\text{P}}_{\text{t}\text{o}\text{t}\text{a}\text{l}}=\text{total pressure}\\ {\text{Χ}}_{\text{A}}\text{= mole fraction of A}\end{array}$.

Answer to Problem 98QAP

0.21 atm.

Explanation of Solution

The balanced chemical equation can be written as follows,

${\text{10 N}}_2\text{O}(g){\text{ + C}}_3{\text{H}}_8(g)\to 10{\text{ N}}_2(g){\text{ + 3 CO}}_2(g)+{\text{ 4H}}_2\text{O}(g)\to (\text{A})$

It is clearly specified in the question that all three main products are in gas state,

If ${\text{N}}_{\text{2}}\text{O}(g)$ behaves ideally in the reservoir, applying the ideal gas law,

$\begin{array}{l}\text{P}\text{V}=\text{n}\text{R}\text{T}\to \text{ (1) where }\\ \text{P=Presssure }\\ \text{V=volume }\\ \text{T=Kelvin Temperature }\\ \text{R=Universal gas constant}\\ \text{and n=number of moles}\end{array}$

$\begin{array}{l}{\text{Before the vessel between the reservoirs were opened, N}}_{\text{2}}\text{O(g) is confined to one reservoir}\text{.Hence;}\\ \text{V=1}\text{.0 L T=450 }\text{C}\text{=(450+273)K=723K}\\ \text{P=1}\text{.4 atm R=0}\text{.08206 (}\frac{\text{L atm}}{\text{K mol}}\text{)}\\ \text{Rearranging equation(1) one can get n = }\frac{\text{PV}}{\text{RT}}\text{®(2)}\\ \text{Substituting the above values to equation(2);}\\ {\text{n}}_{{\text{N}}_{\text{2}}\text{O(g)}}\text{ =}\frac{\text{1}\text{.4 atm x1}\text{.0 L}}{\text{0}\text{.0826(}\frac{\text{L atm}}{\text{K mol}}\text{)×723 K}}\text{ =0}\text{.024 mol(sig}\text{.figures)}\\ {\text{Before the vessel between the reservoirs were opened, C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g) is confined to one reservoir}\text{.Hence;}\\ \text{V=1}\text{.0 L T=450 }\text{C}\text{=(450+273)K=723K}\\ \text{P=1}\text{.2 atm R=0}\text{.08206 (}\frac{\text{L atm}}{\text{K mol}}\text{)}\\ \text{Rearranging equation(1) one can get n = }\frac{\text{PV}}{\text{RT}}\text{®(2)}\\ \text{Substituting the above values to equation(2);}\\ {\text{n}}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g)}}\text{ =}\frac{\text{1}\text{.2 atm x1}\text{.0 L}}{\text{0}\text{.0826(}\frac{\text{L atm}}{\text{K mol}}\text{)×723 K}}\text{ =0}\text{.020 mol(sig}\text{.figures)}\end{array}$

$\begin{array}{l}\text{Hence it is clear that the two reactants are not mixed exactly in the stoichiometric proportions}\text{.}\\ \text{Therefore it is important to find the limiting reagent before proceeding with any calculations}\text{.}\\ \text{According to the stoichiometry of equation (A);}\\ {\text{N}}_{\text{2}}\text{O(g) 10 mol}\equiv {\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g) 1mol}\\ {\text{N}}_{\text{2}}\text{O(g) 0}\text{.024 mol}\to \text{?}\\ {\text{The amount of C}}_{\text{3}}{\text{H}}_{\text{8}}{\text{(g) that reacts with N}}_{\text{2}}\text{O(g) (0}\text{.024 mol)=}\frac{{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g) 1mol}}{{\text{N}}_{\text{2}}\text{O(g) 10 mol}}{\text{×N}}_{\text{2}}\text{O(g) 10 mol=0}{\text{.0024 mol C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g) }\\ {\text{Therefore it is clear that N}}_{\text{2}}\text{O(g) is the limiting reagent under this reaction conditions}\text{.}\\ {\text{The number of C}}_{\text{3}}{\text{H}}_{\text{8}}{\text{(g) that is left in the vessel without reacting with N}}_{\text{2}}\text{O(g) =(0}\text{.020-0}\text{.0024) mol=0}\text{.0176 mol}\\ \text{According to the stoichiometry of equation (A);}\\ {\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g) 1mol}\equiv {\text{N}}_{\text{2}}\text{(g) 10 mol}\\ \text{0}{\text{.0024 mol C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g)}\to \text{?}\\ {\text{The amount of N}}_{\text{2}}\text{(g) formed during the reaction=}\frac{{\text{N}}_{\text{2}}\text{(g) 10 mol}}{{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g) 1mol}}\text{×0}{\text{.0024 mol C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g)=0}{\text{.024 mol N}}_{\text{2}}\text{(g)}\\ {\text{C}}_{\text{3}}{\text{H}}_{\text{8}}{\text{(g) 1molºCO}}_{\text{2}}\text{(g) 3 mol}\end{array}$

$\begin{array}{l}\text{0}{\text{.0024 mol C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g)}\to \text{?}\\ {\text{The amount of N}}_{\text{2}}\text{(g) formed during the reaction=}\frac{{\text{CO}}_{\text{2}}\text{(g) 3 mol}}{{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g) 1mol}}\text{×0}{\text{.0024 mol C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g)=0}{\text{.0072 mol CO}}_{\text{2}}\text{(g)}\\ {\text{C}}_{\text{3}}{\text{H}}_{\text{8}}{\text{(g) 1molºH}}_{\text{2}}\text{O (g) 4 mol}\\ \text{0}{\text{.0024 mol C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g)}\to \text{?}\\ {\text{The amount of N}}_{\text{2}}\text{(g) formed during the reaction=}\frac{{\text{ H}}_{\text{2}}\text{O (g) 4 mol}}{{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g) 1mol}}\text{×0}{\text{.0024 mol C}}_{\text{3}}{\text{H}}_{\text{8}}\text{(g)=0}{\text{.0096 mol H}}_{\text{2}}\text{O (g)}\\ \text{The total number of gas molecules that remain in the reservoir after completion of the reaction}\\ \text{= (0}\text{.0176 + 0}\text{.0096 +0}\text{.0072 +0}\text{.024) mol=0}\text{.0584 mol}\end{array}$

$\begin{array}{l}\text{Now, assuming that all the gas species left in the total reservoirs behaves ideally using the ideal gas law equation(1)}\\ \text{n =0}\text{.0576 mol T=450 }\text{C}\text{=(450+273)K=723K(temperature doesn't change in the process)}\\ \text{R=R=0}\text{.08206 (}\frac{\text{L atm}}{\text{K mol}}\text{) V=2}\text{.0 L}\\ \text{rearranging equation (1) we have}\\ {\text{P}}_{\text{T}}\text{ = }\frac{\text{nRT}}{\text{V}}{\text{ where P}}_{\text{T}}\text{=total pressure in the reaction vessel after reaction is completed®(3)}\\ \text{Substituting the above values to (3)}\\ {\text{P}}_{\text{T}}\text{=}\frac{\text{0}\text{.0584 mol×0}\text{.0826(}\frac{\text{L atm}}{\text{K mol}}\text{)×723 K}}{\text{2}\text{.0 L}}\text{=1}\text{.73 atm(sig}\text{.figures)}\\ {\text{Let partial pressure of CO}}_{\text{2}}{\text{(g) after reaction undergoes completion = P}}_{{\text{CO}}_{\text{2}}\text{(g)}}\\ {\text{P}}_{{\text{CO}}_{\text{2}}\text{(g)}}{\text{=P}}_{\text{T}}{\text{× mole fraction of CO}}_{\text{2}}\text{(g) (QDalton's law of partial pressure)}\\ {\text{P}}_{{\text{CO}}_{\text{2}}\text{(g)}}{\text{=P}}_{\text{T}}\text{×}\frac{{\text{moles of CO}}_{\text{2}}\text{(g) }}{\text{total moles (g) }}\\ {\text{substituting P}}_{\text{T}}\text{=1}{\text{.73 atm and moles of CO}}_{\text{2}}\text{(g) =0}\text{.0072 mol total moles=0}\text{.0584 mol}\\ {\text{P}}_{{\text{CO}}_{\text{2}}\text{(g)}}\text{= 1}\text{.73 atm ×}\frac{\text{0}\text{.0072 mol}}{\text{0}\text{.0584 mol}}\text{=0}\text{.21 atm}\end{array}$.

Interpretation Introduction

(c)

Interpretation:

The partial pressure of CO₂ after the reaction undergoes completion should be calculated.

Concept Introduction:

If there is a mixture of gasses in a container the total pressure exerted is the sum of partial pressure of the gasses present in the container. Partial pressure is defined as the pressure exerted by the gas if the gas was alone in the container.

$\begin{array}{l}{\text{P}}_{\text{t}\text{o}\text{t}\text{a}\text{l}}={\text{ P}}_1{\text{ + P}}_2{\text{ + P}}_3\\ \text{where }{\text{P}}_{\text{t}\text{o}\text{t}\text{a}\text{l}}=\text{total pressure}\\ {\text{P}}_{\text{i}}{\text{ =partal pressure of i}}^{\text{t}\text{h}}\text{ gas component where i=1, 2, 3}\end{array}$

Partial pressure can be derived as follows

$\begin{array}{l}{\text{P}}_{\text{A}}={\text{P}}_{\text{t}\text{o}\text{t}\text{a}\text{l}}\times {\text{Χ}}_{\text{A}}\\ {\text{where P}}_{\text{A}}=\text{partial pressure of }\text{A}\\ {\text{P}}_{\text{t}\text{o}\text{t}\text{a}\text{l}}=\text{total pressure}\\ {\text{Χ}}_{\text{A}}\text{= mole fraction of A}\end{array}$.

Answer to Problem 98QAP

Zero.

Explanation of Solution

Let partial pressure of ${\text{N}}_{\text{2}}\text{O}(g)$ after reaction undergoes completion ${\text{= P}}_{{\text{N}}_{\text{2}}\text{O}(g)}$

$\begin{array}{l}{\text{P}}_{{\text{N}}_{\text{2}}\text{O(g)}}{\text{=P}}_{\text{T}}{\text{× mole fraction of N}}_{\text{2}}\text{O(g) (QDalton's law of partial pressure)}\\ {\text{P}}_{{\text{N}}_{\text{2}}\text{O(g)}}{\text{=P}}_{\text{T}}\text{×}\frac{{\text{moles of N}}_{\text{2}}\text{O(g) }}{\text{total moles (g) }}\end{array}$

${\text{N}}_{\text{2}}\text{O}(g)$ is the limiting agent.

Hence it is clear that there is no ${\text{N}}_{\text{2}}\text{O}(g)$ after the reaction undergoes the completion.

$\Rightarrow {\text{moles of N}}_2\text{O}(g)=0.0000\text{ mol}$

$\begin{array}{l}{\text{substituting P}}_{\text{T}}=1.73\text{ atm and }{\text{moles of N}}_2\text{O}(g)=0.0000\text{ mol total moles=0}\text{.0584 mol}\\ {\text{P}}_{\text{C}{\text{O}}_2(g)}=\text{ 1}\text{.73 atm }\times \frac{0.0000\text{ mol}}{0.0584\text{ mol}}=0\text{ atm}\end{array}$.