An empty 4.00 L steel vessel is filled with 1.00 atm of CH4(g) and 4.00 atm of O2(g) at 300^∘ C

An empty 4.00 L steel vessel is filled with 1.00 atm of...

An empty 4.00 $\mathrm{L}$ steel vessel is filled with 1.00 atm of $\mathrm{CH}_{4}(g)$ and 4.00 $\mathrm{atm}$ of $\mathrm{O}_{2}(g)$ at $300^{\circ} \mathrm{C}$ . A spark causes the $\mathrm{CH}_{4}$ to burn completely, according to the equation: $$\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H^{\circ}=-802 \mathrm{kJ}$$ (a) What mass of $\mathrm{CO}_{2}(g)$ is produced in the reaction? (b) What is the final temperature inside the vessel after combustion, assuming that the steel vessel has a mass of 14.500 $\mathrm{kg}$ , the mixture of gases has an average molar heat capacity of 21 $\mathrm{J} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)$ , and the heat capacity of steel is 0.449 $\mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)$ ? (c) What is the partial pressure of $\mathrm{CO}_{2}(g)$ in the vessel after combustion?

An empty 4.00 $\mathrm{L}$ steel vessel is filled with 1.00 atm of $\mathrm{CH}_{4}(g)$ and 4.00 $\mathrm{atm}$ of $\mathrm{O}_{2}(g)$ at $300^{\circ} \mathrm{C}$ . A spark causes the $\mathrm{CH}_{4}$ to burn completely, according to the equation:
$$\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H^{\circ}=-802 \mathrm{kJ}$$
(a) What mass of $\mathrm{CO}_{2}(g)$ is produced in the reaction?
(b) What is the final temperature inside the vessel after combustion, assuming that the steel vessel has a mass of 14.500 $\mathrm{kg}$ , the mixture of gases has an average molar heat capacity of 21 $\mathrm{J} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)$ , and the heat capacity of steel is 0.449 $\mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)$ ?
(c) What is the partial pressure of $\mathrm{CO}_{2}(g)$ in the vessel after combustion?

Key Concepts

Ideal Gas Law and Partial Pressures

The ideal gas law is a fundamental relation in thermodynamics that connects pressure, volume, temperature, and the number of moles of a gas. It is used to calculate changes in these variables during chemical reactions. Additionally, the concept of partial pressures—where the total pressure of a gas mixture is the sum of the individual pressures of its components—is crucial for understanding the behavior of gases in confined systems after a chemical reaction.

Enthalpy of Reaction

This key concept refers to the heat change that occurs during a chemical reaction at constant pressure, often denoted as ?H. For exothermic reactions like combustion, energy is released into the surroundings. Understanding and applying this concept allows one to relate the energy released during the reaction to other thermodynamic processes, such as heating of the reaction vessel.

Calorimetry and Heat Transfer

Calorimetry involves measuring the heat exchanged during a chemical reaction or physical change. In this context, it combines the concept of specific heat capacity—the energy required to raise the temperature of a substance by one degree—with the conservation of energy principle, accounting for heat absorbed by both the reaction mixture and the container to determine the final temperature.

Reaction Stoichiometry

This concept involves using the balanced chemical equation to determine the quantitative relationships between reactants and products. It provides the mole ratios required to calculate how much of a substance is produced or consumed in a reaction, which is essential for predicting reaction outcomes in chemical processes.

Limiting Reagent Concept

In any chemical reaction where the reactants are not present in the exact stoichiometric ratios, the limiting reagent is the reactant that is completely used up first. This concept is critical because the amount of product formed is determined by the quantity of the limiting reagent, affecting the efficiency and yield of the reaction.

Transcript

00:03 Okay, so for the first part of this question, we're given a 4 -liter container.

00:07 We've got one atmosphere of methane and four atmospheres of oxygen at 300 degrees celsius.

00:13 We get a spark which results in complete combustion, according to the equation shown.

00:21 And the question is what mass of co2 gas is produced in the reaction? that's the question for part a.

00:28 So first thing we can do is we can recognize that the, the partial pressures of these gases are proportional to the amounts of moles, meaning that we have four times as much oxygen as we have methane.

00:44 We only need twice as much oxygen as methane for the combustion to occur.

00:48 So that tells us right off the bat that we're going to run out of methane first.

00:56 So if we want to find out how many moles of co2 we create and then convert that into grams, first, we need to know how many moles of methane we used.

01:07 So if we look at our ideal gas law, pv equals n -r -t, where our pressure is one atmosphere for the ch4.

01:23 We're trying to solve for n.

01:25 So pv over rt equals n.

01:29 So equals one atmosphere times the volume, which is 4 liters, divide by r, since we're in atmospheres, at 0 .0821.

01:43 And the temperature is 300 celsius.

01:48 And we need to convert that into kelvin.

01:50 You always need to be in kelvin.

01:52 So that's going to be 573 kelvin.

01:58 And so if we multiply that out, we get four times one, divide by .0821.

02:08 And then divide also by 573.

02:13 0 .085 moles of ch4.

02:19 Now for every one mole of ch4, we generate one mole of co2, which means that that is also the same as the number of moles of co2 that we create.

02:33 Now the question is how many grams of co2 do we create? and so we take our 0 .085 and multiply it by the molar mass of co2.

02:44 Carbon is 12, oxygen is 16, there are two of them, so that's 32, 12 and 32 is 44 grams per mole, and that gives us 3 .74 grams of co2.

03:07 So that's part a.

03:13 Part b is going to require us to do some thermochemistry.

03:21 So the first thing that we need to do is we know that we have, so that each mole is giving us 802 kilojoules of energy.

03:37 So 802 kilojoules per mole is the energy that we get from combusting our methane.

03:52 And then we need to multiply that by the number of moles of methane that we combusted.

03:59 Which was, if you recall, 0 .085.

04:11 To get the total number of kilojoules, 68 .17 kilojoules or 68 ,170 joules, might be a little bit more useful to us.

04:33 And then what we're going to do is we're going to say that our heat capacity for our steel, we have 14 .5 kilograms and 0 .49 jules per gram degree celsius.

04:56 So 14 ,500 grams multiply by 0 .449 joules per gram degree celsius gives us 6 ,510 .5 joules per degree celsius...

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