Consider the following reaction: CH4(g)+2 O2(g) ⟶CO2(g)+2 H2 O(l) ΔH=-891 kJ Calculate the en

Consider the following reaction: CH4(g)+2 O2(g) ⟶CO2(g)+2 H2...

Consider the following reaction: $\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-891 \mathrm{~kJ}$ Calculate the enthalpy change for each of the following cases: a. $1.00 \mathrm{~g}$ methane is burned in excess oxygen. b. $1.00 \times 10^{3} \mathrm{~L}$ methane gas at 740 . torr and $25^{\circ} \mathrm{C}$ is burned in excess oxygen.

Consider the following reaction:
$\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-891 \mathrm{~kJ}$
Calculate the enthalpy change for each of the following cases:
a. $1.00 \mathrm{~g}$ methane is burned in excess oxygen.
b. $1.00 \times 10^{3} \mathrm{~L}$ methane gas at 740 . torr and $25^{\circ} \mathrm{C}$ is burned in excess oxygen.

Key Concepts

Mole Concept

The mole is a fundamental unit in chemistry that represents a specific number of particles, typically atoms or molecules, and is critical for converting between mass and the number of particles. By using molar mass, one can determine how many moles are present in a given mass, which then allows for quantitative analysis in reactions.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It uses the mole ratios from the balanced chemical equation to relate the amounts of each substance involved, which is essential when calculating the energy changes associated with a reaction based on the specific quantity of a reactant burned.

Enthalpy Change

Enthalpy change refers to the heat absorbed or released during a chemical reaction at constant pressure. For combustion reactions, this change is often exothermic, meaning energy is released. The enthalpy change per mole of a reactant is used to compute the overall heat effect when a certain quantity of the reactant is consumed.

Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance combines with oxygen to form products, typically releasing heat in the process. These reactions frequently involve hydrocarbons and oxygen, and the heat released (enthalpy of combustion) can be used as a measure of the energy content of the fuel.

Ideal Gas Law

The ideal gas law, expressed as PV = nRT, relates the pressure, volume, temperature, and number of moles of a gas. This law is particularly useful for converting a given volume of gas at known pressure and temperature to moles, which is an important step when calculating reaction energetics involving gaseous reactants.

Transcript

00:01 In this video, we're going to review how to find the inbuilt they changed based on a given reaction equation and the delta h.

00:12 So we have methane, which is a ch bore, and it's in.

00:15 The presence of oxygen is being burned to give us carbon dioxide and two moles of water.

00:22 And the given delta h is negative 8 91 kill jules.

00:28 I know this is excellent, thermic, as most conversion reactions are, so he is going to be a product on the right hand side.

00:36 So i added the plus 891 killer jewels over here on the right.

00:43 So the first question asks in part a if we're given one gram of methane, how much your words are impossibly impossibly change here in killer jewels.

00:53 So this would be a pretty much basic story geometric process.

00:58 We would take one gram of methane and we would figure out how animals have nothing we have, and we can just you know, what? our molar masses from the periodic table i would have had an added that up.

01:09 That is 16 grams and one more in one more.

01:20 And we know that one more of methane in that equation when well, of ch war generates that 891 killer jules case.

01:31 We pulled that from they this racial here from the equation.

01:37 I know one mole of ch four will generate 100 91 killer jewel's worth of energy.

01:44 And then from there we can just go ahead and cancel out our units, and we should end up with a result of 55.6875 i would keep it simple and round it to 55.7 killed.

02:05 Jules, we're and they'll be.

02:11 That should be negative.

02:13 Where the second part were given a volume of one times 10 to the third leaders a pressure of 740 tour and were given a temperature as well 25 degrees celsius.

02:25 That's room temperature.

02:26 I went ahead and added the 273 degrees to get calvin.

02:32 We always need to work with calvin when we're working with gas laws in this particular case.

02:38 So i see volume, pressure and temperature.

02:41 I know that i can use pv equals an arty and this is for a given amount of oxygen gas over here.

02:49 Wrong.

02:50 The on the right.

02:52 We have oxygen gas here.

02:54 So using pv equals in our tea and notice.

02:58 I did switch the 740 tour overto atmospheres...