Chapter 14, Problem 94A

(a)

Interpretation Introduction

Interpretation-To determine the partial pressure of $\text{NO}$ in the sealed container.

Introduction- Ideal Gas law: pressure, volume, and temperature can be related as:

  $PV=nRT$

Where, P = pressure in $atm$ V= volume in $L$ T = temperature in $K$ n = mole

R = universal gas constant = $0.08206\text{atm}\cdot \text{L}/\text{mol}\cdot \text{K}$

Limiting Reactant -It is the reactants that finish first when the reaction stops and limits the amount of product formation.

Answer to Problem 94A

The partial pressure of $NO$is = $1.61\text{ atm}$

Explanation of Solution

Given:

Mass of $N{H}_3$ = $34.0\text{g}$

Mass of $O_2$ = $96.0\text{g}$

Molar mass of $N{H}_3$ = $17.0\text{g}/\text{mol}$

Molar mass of $O_2$ = $32\text{g}$

Mole ratio of $\text{NO}$ : ${\text{NH}}_{\text{3}}$ = $4:4$

Mole ratio of $\text{NO}$ : ${\text{O}}_{\text{2}}$ = $4:5$

1. Limiting reactant can be found as:

a. Amount of $NO$ produced by $30.0gN{H}_3$ can be calculated as:

  $34.0gN{H}_3\times \frac{1molN{H}_3}{17.0gN{H}_3}\times \frac{4molNO}{4molN{H}_3}=2molNO$

b. Amount of produced by can be calculated as:

  $96g{O}_2\times \frac{1mol{O}_2}{32g{O}_2}\times \frac{4molNO}{5mol{O}_2}=2.4molNO$

Hence, the limiting reactant is ammonia as it produced a lesser amount of product.

The theoretical yield of $NO$ will be $2molNO$

2. Partial pressure can be calculated using the ideal gas law as:

Given:

Volume = $40.0L$

Temperature = ${120}^{\circ }C+273K=393K$

Mole = $2molNO$

  $\begin{array}{l}PV=nRT\\ P\times 40.0L=2.0mol\times 0.08206\frac{atm*L}{mole*K}\times 393K\\ P\times 40.0L=56.8atm*L\\ P=1.61atm\end{array}$

Therefore, the partial pressure of $NO$ is= $1.61atm$

Conclusion

Therefore, the partial pressure of $NO$ is= $1.61atm$

(b)

Interpretation Introduction

Interpretation- To determine the total pressure

Introduction-

When the reactant stops, product and excess reactants remain.

Excess reactant is the reactant that remains when the reaction stops. The amount of excess reactant can be calculated using stoichiometry.

The total pressure can be calculated using Dalton’s Law of partial pressure:

  $P_{total}=P_1+P_2+P_3\mathrm{........}$

Where P total =total pressure

P₁,P_2,P₃ are partial pressure of gases.

Answer to Problem 94A

Therefore, the total pressure in the container = 4.43atm

Explanation of Solution

Oxygen gas is an excess reactant as ammonia is calculated as limiting. So, oxygen gas, nitrogen monoxide, and water remain when the reaction stops.

Amount of oxygen gas reacted:

  $\begin{array}{l}34.0gN{H}_3\times \frac{1molN{H}_3}{17.0gN{H}_3}\times \frac{5mol{O}_2}{4molN{H}_3}=2.5mol{O}_2\\ Totalmoleof{O}_{2}present=96g{O}_2\times \frac{1mol{O}_2}{32g{O}_2}=3mol{O}_2\\ Moleof{O}_{2}remain=3mol-2.5mol=0.5mol\end{array}$

Partial pressure of oxygen gas can be calculated as:

  $\begin{array}{l}PV=nRT\\ P\times 40.0L=0.5mol\times 0.08206\frac{atm\cdot L}{mole\cdot K}\times 393K\\ P\times 40.0L=16.12atm\cdot L\\ P=0.40atm\end{array}$

Amount of water produced:

  $34.0gN{H}_3\times \frac{1molN{H}_3}{17.0gN{H}_3}\times \frac{6mol{H}_{2}O}{4molN{H}_3}=3mol{H}_{2}O$

Partial pressure of water

  $\begin{array}{l}PV=nRT\\ P\times 40.0L=3.0mol\times 0.08206\frac{atm\cdot L}{mole\cdot K}\times 393K\\ P\times 40.0L=96.74atm\cdot L\\ P=2.42atm\end{array}$

Total pressure can be calculated as:

  $\begin{array}{l}{P}_{total}=P_1+P_2+P_3\mathrm{........}\\ P_{total}=1.61atm+0.40atm+2.42atm=4.43atm\end{array}$