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AP Chemistry : Other Gas Concepts

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Example Questions

Example Question #2 : Gases

COCl₂ (g)⇌ CO (g) + Cl₂ (g)

Initially a reaction chamber contains only 2 moles of COCl₂, after the reaction has reached the equilibrium shown above the pressure is measures to be 2.60 atm. What is the most likely cause for the increase in pressure?

Possible Answers:

Heat is consumed during the course of the reaction

CO is larger than COCl₂

CL₂ is a diatomic gas

There are more molecules in the chamber, increasing the liklihood of atomic collisions with the wall of the container leading to increased pressure.

Heat is released during the course of the reaction

Correct answer:

There are more molecules in the chamber, increasing the liklihood of atomic collisions with the wall of the container leading to increased pressure.

Explanation:

The equilibrium expression indicates that one mole of COCl₂(g) will decompose into CO (g) and Cl₂ (g). One mole of gas is present on the left side of the equation, and two moles of gas are present on the right side of the equation. The increases in pressure is due to the increase in the number of gas molecules in the reaction chamber.

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Example Question #1 : Other Gas Concepts

A gas is at STP

What does STP refer to?

Possible Answers:

0 C, 1 atm

0 C, 25 atm

25 C, 25 atm

25 C, 1 atm

298 K, 25 atm

Correct answer:

0 C, 1 atm

Explanation:

Standard temperature and pressure is 0⁰C and 1 atm

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Example Question #2 : Other Gas Concepts

What is the molar mass of a gas with a density of $\small 3.21\frac{g}{L}$ at STP?

$\small R=0.08206\frac{L*atm}{mol*K}$

Possible Answers:

$71.9\frac{g}{mol}$

$\small 45.8\frac{g}{mol}$

$\small 36.0\frac{g}{mol}$

$\small \small 98.3\frac{g}{mol}$

Correct answer:

$71.9\frac{g}{mol}$

Explanation:

Since we have the density of the mystery gas, we can rearrange the ideal gas law so that the remaining factors are equal to the density of the gas.

PV=nRT

We can redefine moles of gas as mass over molar mass.

$n=\frac{mol}{molar\ mass}=\frac{m}{M}$

$\small PV = \frac{m}{M}RT$

We can now rearrange the equation to solve for density (mass per volume), which is given in the question.

$\small \frac{m}{V} = \frac{PMM}{RT} = 3.21\frac{g}{L}$

Using this set up and the values for standard temperature and pressure, we can solve for the molar mass.

$T_{STP}=273K$

$P_{STP}=1atm$

$\small 3.21\frac{g}{L} = \frac{(1atm)M}{(0.08206)(273K)}$

$\small M = 71.9\frac{g}{mol}$

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Example Question #13 : Gases

Which of the following will not be found in the atmosphere for an extended period of time?

Possible Answers:

N_2

O_2

O_3

H_2

Correct answer:

Explanation:

Most gases exist as diatomic molecules in nature. O_2 , N_2 , and H_2 are all common gases found in the atmosphere and the air we breathe.

Ozone, O_3 , is a rare exception to the diatomic bonding principle, and is also found in significant quantities in the atmosphere.

Monatomic oxygen, , has only six valence electrons and is relatively unstable; this compound would not be found naturally in the atmosphere.

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Example Question #11 : Gases

A flexible box contains four moles of octane gas at STP with an excess of oxygen. If the octane is combusted according to the following reaction and the box is allowed to expand, what is the change in volume?

$2C_8H_{18}_{(g)} + 25O_{2(g)}\rightarrow16CO_{2(g)}+18H_2O_(g)$

Possible Answers:

The size of the box increases by 13.5L

More information is needed in order to solve

The size of the box increases by 68L

The size of the box increases by 315L

Correct answer:

The size of the box increases by 315L

Explanation:

This question does not acually deal with the ideal gas law, but only with the standard volume of a gas at STP. At standard temperature and pressure, one mole of gas has a volume of 22.5L.

We know that initially we have four moles of octane and excess oxygen. This means that octane will be the limiting reagent for the combustion reaction, since oxygen is in excess.

The total moles of reactant gas when utilizing all four moles of octane is 54, while the total moles of product gas is 68.

$2(2C_8H_{18}_{(g)} + 25O_{2(g)}\rightarrow16CO_{2(g)}+18H_2O_(g))$

$4C_8H_{18}_{(g)} + 50O_{2(g)}\rightarrow32CO_{2(g)}+36H_2O_(g)$

There is a total change in moles of:

68-54=14

Since we end the reaction with 14 more moles of gas than we started, the volume must increase:

$14mol*\frac{22.5L}{1mol}=315L$

The change in volume is 315L .

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Example Question #13 : Gases

What is the osmotic pressure exerted by a solution of 2M CaCl₂at room temperature? (R=0.082 L*atm/K/mol)

Possible Answers:

147 atm

18 atm

6 atm

73 atm

250 atm

Correct answer:

147 atm

Explanation:

$\pi=iMRT$

The van't Hoff factor for CaCl₂ is 3 since the molecule dissociates in solution to 3 ions. T is in absolute temperature (25+273=298). This makes the osmotic pressure:

3*2M*0.082*298.

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Example Question #331 : Ap Chemistry

At $25.0^{0}C$ the vapor pressure of pure water is 23.8 mmHg . What is the minimum amount of liquid water we need to put in a $10.0\: L$ flask to achieve this vapor pressure. $R\: = 0.08206 \frac{L\: atm}{mol K}$

Possible Answers:

It does not matter the amount of water, the vapor pressure will always be reached

$0.231\: L$

$1.28 \times 10^{-2}\: mol$ which is equivalent to $0.231\: g$

$23.8\: g$

Correct answer:

$1.28 \times 10^{-2}\: mol$ which is equivalent to $0.231\: g$

Explanation:

We use the ideal gases equation to know the mol of water we need in the gas phase to generate a pressure of $23.8\: mmHg$ in a $10.0\: L$ flask at $25.0^{0}C$ .

$n= \frac{10.0\: L\: \frac{23.8}{760}\: atm}{0.08206\frac{L\: atm}{mol\: K}\left ( 25.0+273 \right )^{0}C}\: =\: 1.28\times 10^{-2}\:mol$

And the mass is:

$1.28\times 10^{-2}\: mol\times 18.01\frac{g}{mol}= 0.231\: g$

If we have less than that amount of water the liquid will evaporate completely but there will not be enough molecules to reach the vapor pressure. More than $0.231\: g$ of water will allow to reach the vapor pressure and there will be also a liquid leftover. Remember that the vapor pressure of a liquid at a fixed temperature is a constant.

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Example Question #13 : Gases

$3.05\: g$ of $NH_{4}NO_{3}$ are introduced in a $2.18\: L$ flask. A vacuum is applied to the flask and its temperature raised to $250^{0}C$ . At this temperature ammonium nitrate decomposes according to the reaction:

$NH_{4}NO_{3(s)}\: \rightarrow\: N_{2}O_{(g)}\: +\: 2H_{2}O_{(g)}$

If the percent yield of this reaction is $80\%$ , which will be the total pressure of the flask after the reaction concludes. $R=0.08206\frac{L\: atm}{mol\: K}$

Possible Answers:

$0.114\: mol$

$0.60\: atm$

$2.25\: atm$

$1.8\: atm$

Correct answer:

$1.8\: atm$

Explanation:

Using the molecular mass we calculate the number of mol of $NH_{4}NO_{3}$ :

$mol\: of \: NH_{4}NO_{3}= \frac{3.05\: g\: NH_{4}NO_{3}}{80.05\: \frac{g}{mol}}=3.81\times 10^{-2} mol \: NH_{4}NO_{3}$

At this moment we could apply the percentage yield, however let's live it for the end. Each mole of $NH_{4}NO_{3}$ decomposed will generate $3 \: mol$ of gas. Then the total pressure assuming $100\%$ yield will be:

$P_{total}=\frac{3\: mol\: of\: NH_{4}NO_{3}\times R\timesT\times T }{V}$

$P_{total}=\frac{3\times 3.81\times 10^{-2}\: mol\: of NH_{4}NO_{3}\times 0.08206\frac{L\: atm}{mol\: K}\times \left ( 250+273 \right )K}{2.18\: L}=$

$=2.25\: atm$

Since only $80\%$ of the $NH_{4}NO_{3}$ decomposes, the real total pressure will be:

$P_{total}=2.25\: atm\times \frac{80}{100}=1.8\: atm$

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Example Question #17 : Gases

It can be argued that one of the greatest innovations in automobile safety is the airbag. A typical driver's side airbag expands to a volume of between 50-60L. For this problem use 55.00L . In older airbag models sodium azide is employed to produce nitrogen gas as shown in the following decomposition reaction:

$2NaN_3_{(s)}\xrightarrow[]{300^{\circ}C} 2Na_{(s)} + 3N_2_{(g)}$

Two more reactions are employed to remove the reactive sodium metal, shift the above reaction equilibrium to the right, and produce more nitrogen gas; however for this problem assume all gas is generated from the first reaction as shown above and that the reaction proceeds to completion.

The gauge pressure is 0.34atm . This is the pressure at which the airbag is deployed. Note: gauge pressure is defined as the additional pressure in the system relative to atmospheric pressure. In this case use STP pressure as atmospheric pressure.

(It is worth noting that this pressure is the initial pressure to fill the airbag, which would provide a rock-hard cushion; therefore in the real world the air bag immediately begins to deflate so as to provide proper cushioning; that fact is not important for solving this problem).

$R= 0.082057 \frac{L\cdot atm}{mol\cdot K}$

What is the minimum mass of $NaN_3_{(s)}$ required to inflate the airbag given the above conditions?

$R= 0.082057 \frac{L\cdot atm}{mol\cdot K}$

Possible Answers:

22.05g

102.07g

204.2g

68.04g

153.10g

Correct answer:

68.04g

Explanation:

First, identify the necessary parameters.

We must use the ideal gas equation:

PV=nRT

$R= 0.082057 \frac{L\cdot atm}{mol\cdot K}$

$T=300^{\circ}C$ and since the molar gas constant given uses Kelvin in it's definition, we must convert $300^{\circ}C$ to Kelvin as follows: 300+ 273= 573K

$P_{(gauge)}= 0.34atm$

Since the pressure given is gauge pressure, we must convert to the actual pressure as follow: $P=P_{gauge} +P_{atmospheric}$

So therefore the pressure we want to use is P=0.34atm + 1atm = 1.34atm

V=55.00L

$2NaN_3_{(s)}\xrightarrow[]{300^{\circ}C} 2Na_{(s)} + 3N_2_{(g)}$

Now that we have listed the given quantities we can see that in order to get the mass of $NaN_3_{(s)}$ needed, we must first find the number of moles of $N_2_{(g)}$ needed to fill the volume of the airbag, then use the molar ratios from the above chemical equation to convert moles of $N_2_{(g)}$ to moles of $NaN_3_{(s)}$ , after which we can use the molar mass of $NaN_3_{(s)}$ to calculate the grams of $NaN_3_{(s)}$ needed to inflate the airbag.

In order to calculate the number of moles of $N_2_{(g)}$ we can employ the Ideal gas law: PV=nRT to find n (the number of moles of gas; which in this case the only gas involved is $N_2_{(g)}$

Plug in the given quantities that have been converted into the correct units and quantities:

$(1.34atm)(55L)=(n)(0.082057 \frac{L\cdot atm}{mol\cdot K}) (573K)$

Solve.

$(73.7atm\cdot L)=(n)(47.019\frac{L\cdot atm}{mole})$

n=1.57moles $N_2_{(g)}$

Now that we know the moles of $N_2_{(g)}$ needed we can use stoichiometry, and the information in the following equation:

$2NaN_3_{(s)}\xrightarrow[]{300^{\circ}C} 2Na_{(s)} + 3N_2_{(g)}$

To determine the number of moles of $NaN_3_{(s)}$ required and then convert the moles of $NaN_3_{(s)}$ to the grams of $NaN_3_{(s)}$ using the molar mass of $NaN_3_{(s)}$ . The molar mass of $NaN_3_{(s)}$ can be found by using a periodic table and adding the molar masses of the constituent elements as shown below:

$Na= 22.990 \frac{g}{mol}$

$N= 14.007\frac{g}{mol}$

Molar mass of $NaN_3= 22.990 {\frac{g}{mole}} + (3)(14.007\frac{g}{mol})= 65.011\frac{g}{mol}$

Given that we calculated the moles of $N_2_{(g)}$ to fill the volume of the airbag to be: n=1.57moles $N_2_{(g)}$ , and the molar ratios from the chemical equation we can set up the final steps of the calculation as shown below:

$NaN_3_{(s)}=1.57molN_2_{(g)}*\frac{2molNaN_3_{(s)}}_{3molN_2_{(g)}\frac}$ $*\frac{65.011gNaN_3_{(s)}}{1molNaN_3_{(s)}}= 68.04g NaN_3_{(s)}$

Therefore in order to inflate this driver's side airbag it would need to contain at least 68.04g of $NaN_3_{(s)}$

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Example Question #12 : Solutions And States Of Matter

Suppose that a container contains $7.5\: grams$ of a gaseous sample of unknown hydrocarbon at STP. If this sample occupies a volume of $5.6\:L$ , what is the identity of the hydrocarbon?

Possible Answers:

$C_{2}H_{2}$

$CH_{3}$

$C_{2}H_{4}$

$C_{2}H_{6}$

Correct answer:

$C_{2}H_{6}$

Explanation:

In this question, we're given the mass and volume of an unknown hydrocarbon filling a container. We're asked to determine the identity of this compound.

To solve this problem, it's important to realize that for any ideal gas at STP (standard temperature and pressure), $22.4\:L$ of the gas will equate to $1\:mol$ of that gas. For this question, we're told that there is $5.6\:L$ present.

$\frac{22.4\: L}{5.6\: L}=\frac{1\: mol}{x}$

$x=0.25\: mol$

Now that we know how many moles of gas are in the container, we can use this information, together with the mass provided to us in the question stem, to determine the molecular mass of the unknown compound.

$\frac{7.5\:grams}{0.25\: moles}=\frac{30\: grams}{mol}$

Looking at the answer choices, the only hydrocarbon that matches this molecular mass is ethane, $C_{2}H_{6}$ .

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AP Chemistry : Other Gas Concepts

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #2 : Gases

Example Question #1 : Other Gas Concepts

Example Question #2 : Other Gas Concepts

Example Question #13 : Gases

Example Question #11 : Gases

Example Question #13 : Gases

Example Question #331 : Ap Chemistry

Example Question #13 : Gases

Example Question #17 : Gases

First, identify the necessary parameters.

$R= 0.082057 \frac{L\cdot atm}{mol\cdot K}$

Example Question #12 : Solutions And States Of Matter

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