10.3: Gas Phase Reactions

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Introduction

Now that we have covered the ideal gas law we can revisit the stoichiometric equations of Chapter 4 and account for gas phase reactants and products in a more meaningful manner. While going over this material you may wish to review the following sections.

3.2: Types of Chemical Reactions (combustion and many types of single displacement involve gasses)
3.5.4: Acid Reactions Evolving Gases (Strong acid reactions with salts of oxyanions [acid anhydrides])
4.1: Stoichiometry
4.2: Limiting & Excess Reagents

In the previous sections we would report gasses in terms of their moles, or even their mass. Lets take a second look at Exercise 4.2.1, where we calculated the mass of carbon dioxide formed upon the combustion of 25.00g glucose with 40.0 g of oxygen. Sure we could solve the problem, but at standard conditions oxygen or carbon dioxide are gasses not solids, and so you would not be able to directly measure their mass, and a more realistic problem would be to define them in terms of measurable quantities like pressure, temperature and volume.

Overview of Stoichiometry:

In stoichiometric calculations you typically have three steps.

Convert measured quantities to moles (assuming you don't start with moles, but something measurable)
Use balanced equations to relate moles of starting material to moles of desired substance sought.
Convert moles of desired substance to measurable quantities (assuming you don't want moles, but something measurable)

mass	solute in solution
n=\(\frac{m}{fw}\)	n=MV	n=\(\frac{PV}{RT}\)
m=mass fw=formula weight	M=Molarity=\(\frac{mol_{solute}}{liter_{solution}}\) V=Volume in liters	P=Pressure V=Volume R=Ideal Gas Constant T=Temperature in Kelvin

As in Chapter 4, it is a good idea to write out and balance the equation, write under it what is given, and then solve the problem. The following problems involves calculations involving all three phases.

Example \(\PageIndex{1}\)

What volume of gas would be produced at 760.0 torr and 298 K if 1.00 g of magnesium metal is dropped into 50.0 ml of 1.00 M Hydrochloric Acid?

Solution
(See Video \(\PageIndex{1}\) for solution)

Step 1: Write out and balance equation and write underneath it what is given:

\[\underset{m=1.00g \\fw= 24.3\frac{g}{mol}}{Mg(s)} \; + \; \underset{0.0500L \\ 1.00 M}{2HCl(aq)} \; \rightarrow \; Mg(Cl)_2(aq)+\underset{P=760.0 atm \\ T=298K \\n=? \\ V=??}{H_2(g)}\]

NOTE: We want \(V_{H_2}\), but have two unknowns (V and n), so we need to calculate n

Step 2: Calculate theoretical yield of n (based on complete consumption of limiting reagent)

\[1.00g \; Mg\left ( \frac{mol \; Mg}{24.3g} \right )\left ( \frac{1}{1mol \; Mg} \right )=0.0411 \\ 0.0500L\left ( \frac{1.00mol \;HCl}{L} \right )\left ( \frac{1}{2mol \;HCl} \right )=.025
\\HCl =limiting \; reagent \\n_{H_2} =0.025\left ( 1mol \; H_2(g) \right )=0.025 mol\;H_2(g)\]

Step 3: Now that we have \(n_{H_2}\), use ideal gas law to solve for V at the conditions of the problem.

\[V=\frac{nRT}{P}=\frac{0.02500mol\left (0.08206\left ( \frac{l \cdot atm}{mol \cdot K} \right ) \right )298K}{760 \; torr\left ( \frac{1atm}{760 \;torr} \right )} = .6113L=611mL\]

Video \(\PageIndex{1}\): 5'35" YouTube showing solution for problem above

Exercise \(\PageIndex{2}\)

Solve the following problems

5.00 g of Mg is added to 50.0 mL of 0.800M HCl. A single displacement reaction occurs in a sealed container at 25.0^oC. What is the pressure of the hydrogen gas generated if the volume above the solution is 100.0 ml, and we ignore any pressure due to the evaporated water?
Nitrogen dioxide is formed in a closed container at 90^oC and 1.00 atm when 1.50 g NO and 2.00 mole of O₂ are mixed. After the reaction is finished the pressure changes to 3.90 atm. What is the final temperature?

Answer a: \[P_{H_{2}}=\frac{n_{H_{2}}RT}{V}=\frac{0.0200mol~H_{2}\left ( 0.08206\frac{L\cdot atm}{mol\cdot K} \right )298K}{0.100L}=4.89atm\]
Answer b: \[T_{2}=T_{1}\left ( \frac{n_{1}}{n_{2}} \right )\left ( \frac{P_{2}}{P_{1}} \right )=363.15K\left ( \frac{2.05mol}{2.025mol} \right )\left ( \frac{3.90atm}{1.00atm} \right )=1433.77K =1400K\]

Contributors and Attributions

Bob Belford (UALR) and November Palmer (UALR)