How much heat, in kilojoules, is evolved in the complete...

How much heat, in kilojoules, is evolved in the complete combustion of (a) $1.325 \mathrm{gC}_{4} \mathrm{H}_{10}(\mathrm{g})$ at $25^{\circ} \mathrm{C}$ and 1 atm; (b) $28.4 \mathrm{LC}_{4} \mathrm{H}_{10}(\mathrm{g})$ at STP; (c) $12.6 \mathrm{L} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})$ at $23.6^{\circ} \mathrm{C}$ and $738 \mathrm{mmHg}$ ?Assume that the enthalpy of reaction does not change significantly with temperature or pressure. The complete combustion of butane, $\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}),$ is represented by the equation $\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})+\frac{13}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+5 \mathrm{H}_{2} \mathrm{O}(1)$ $$\Delta_{\mathrm{r}} H^{\circ}=-2877 \mathrm{kJ} \mathrm{mol}^{-1}$$

How much heat, in kilojoules, is evolved in the complete combustion of (a) $1.325 \mathrm{gC}_{4} \mathrm{H}_{10}(\mathrm{g})$ at $25^{\circ} \mathrm{C}$ and 1 atm; (b) $28.4 \mathrm{LC}_{4} \mathrm{H}_{10}(\mathrm{g})$ at STP;
(c) $12.6 \mathrm{L} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})$ at $23.6^{\circ} \mathrm{C}$ and $738 \mathrm{mmHg}$ ?Assume that the enthalpy of reaction does not change significantly with temperature or pressure. The complete combustion of butane, $\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}),$ is represented by the equation $\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})+\frac{13}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+5 \mathrm{H}_{2} \mathrm{O}(1)$
$$\Delta_{\mathrm{r}} H^{\circ}=-2877 \mathrm{kJ} \mathrm{mol}^{-1}$$

Key Concepts

Ideal Gas Law

The ideal gas law (PV = nRT) is used to relate the pressure, volume, temperature, and number of moles of a gas. This law is essential when dealing with gases under various conditions, especially when converting gas volumes at non-standard conditions to moles. It provides a reliable method for calculating the amount of gas present based on measured parameters.

Molar Mass and Mole Calculations

Converting a measured mass of a substance to moles using its molar mass is a fundamental step in quantitative chemistry. This allows for the application of stoichiometric relationships from the balanced equation to determine the corresponding reaction enthalpy, showing the connection between the mass of a fuel and the energy released during combustion.

Standard and Non-Standard Conditions

Understanding the difference between standard conditions (such as STP, where temperature and pressure are defined) and non-standard conditions is important for accurate calculations. When gas volumes are provided at conditions other than STP, adjustments using the ideal gas law or the concept of molar volume are required to correctly determine the number of moles, ensuring that further calculations, such as energy evolution, are accurate.

Enthalpy of Reaction

This concept refers to the heat change that occurs during a chemical reaction at standard conditions. In the context of combustion reactions, the standard enthalpy of combustion (?rH) represents the energy released when one mole of a substance completely reacts with oxygen to form combustion products. It is fundamental for calculating the heat evolved in a reaction based on the number of moles of the reactant consumed.

Reaction Stoichiometry

Reaction stoichiometry involves using the balanced chemical equation to determine the proportions in which reactants combine and products form. In combustion calculations, stoichiometric coefficients indicate how many moles of oxygen are required and how many moles of combustion products result from a given amount of fuel, which is essential for relating the amount of substance to the energy released.

Transcript

00:01 So for this question, we are going to be using this balanced chemical equation for the combustion of butane.

00:07 And for this reaction, they tell us that the enthalpy of this reaction is equal to negative 2 ,87 kilojoules per mole.

00:19 For the first part of this question, they are wanting us to find the heat evolved in kilojoules from this reaction with a certain mass of butane.

00:27 So the mass of butane that they give us is equal to 1 .325 grams of butane.

00:38 We are looking for the heat evolved in kilograms, so that's what we're looking for.

00:43 They tell us that the pressure for this reaction is equal to one atmosphere, and they tell us that the temperature for this reaction is equal to 25 degrees celsius.

00:56 I'm just going to add 273 kelvin to this.

01:01 So our temperature is going to be 298 kelvin.

01:06 We also know that the molar mass of butane is equal to 58 .12 grams per mole.

01:15 All right.

01:16 So to first start out this question, we are looking for the heat evolved.

01:20 But before we can find the heat evolved, we need to first find out how many moles of butane we have.

01:25 So our moles of butane so since we have the mass and we have the molar mass we can just take our mass of butane that we have and divide that by our molar mass so 58 .12 grams per mole so our moles of butane is 0 .02280 moles of butane is 0 .022280 moles of butane so for this reaction we know that for every one mole of c butane that we have we give off 2877 kilojoules of heat so for one mole of butane that's equal to negative 2 ,87 kilojoules of heat so let's find the heat associated with this reaction, so heat associated.

02:49 So to find heat associated, we're just going to take our moles of butane that we have, and we are going to times that by negative 2 ,87 kilojoules, and divide that by one mole of butane.

03:11 So when we calculate this out, our heat associated is going to be negative 65 .59 kilojoules.

03:23 So this is heat associated.

03:25 We want heat evolved.

03:26 So to find heat evolved, we're just going to take the negative sign off.

03:29 So 65 .59 kilojoules.

03:34 So this would be the heat evolved for the first part of this problem.

03:39 For the second part of this problem, they give us a volume of butane...

Please give Ace some feedback