Calculate the enthalpy change when 1.00 g of methane is...

Calculate the enthalpy change when 1.00 g of methane is burned in excess oxygen according to the reaction
$$\begin{array}{c}{\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)} \\ {\Delta H=-891 \mathrm{kJ} / \mathrm{mol}}\end{array}$$

Key Concepts

Enthalpy Change

Enthalpy change, often represented as ?H, is the measure of heat absorbed or released during a chemical reaction at constant pressure. It indicates whether the reaction is exothermic (releases heat) or endothermic (absorbs heat) and is vital for understanding energy flow in chemical processes.

Combustion Reactions

Combustion reactions involve a substance reacting with oxygen to form products, typically releasing energy in the form of heat and light. They are common examples in thermodynamics, illustrating the concepts of exothermic reactions and energy generation.

Mole and Molar Mass

The mole is a fundamental unit in chemistry used to count particles, and the molar mass is the mass of one mole of a substance. These concepts are essential for converting between the mass of a substance and the number of moles involved in a chemical reaction, allowing for accurate energy calculations based on stoichiometry.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It uses the balanced chemical equation to determine the proportions of substances, which is crucial when calculating how much energy is released or absorbed per given amount of a reactant.

Transcript

00:01 So this question gives us the chemical reaction for burning methane with molecular oxygen to form carbon dioxide and liquid water.

00:07 It asks us to find the overall enthalpy h given the change of enthalpy delta h as negative 891 kilojoules per mole, as well as the mass of methane involved, which is 1 gram.

00:21 So before we can begin, this chemical reaction isn't balanced.

00:23 And you should always make sure they are before you start, so your math is correct.

00:26 So you have one carbon on the left side and one carbon on the right side.

00:30 Which checks out, but we have four hydrogen on the left side and only two hydrogen on the right side.

00:36 We also have four oxygen on the left side and only three oxygen total on the right side.

00:42 So pretty quickly we can see that doubling the number of liquid water on the right side will both double the number of hydrogen involved on the right side, as well as add one oxygen, meaning that both are the same as the left side now and our chemical reaction is balanced.

00:57 So we know our math is correct moving forward.

00:59 We also know from the question that we have excess oxygen involved, meaning that we know our limited reagent is the methane, and it's what we'll use to calculate the overall enthalpy, which is calculated by multiplying the change in enthalpy delta h by the mass.

01:14 However, these factors don't have the same units currently as the change in entropy is in moles and the mass is in grams, so we have to convert our mass into moles using the molecular mass of methane.

01:25 So we'll take our mass of 1 gram of methane, multiplied by the molar mass, which we find using the overall mass of the constituent elements of methane...

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