Question

Consider the following reaction: \(\mathrm{CH}_{3} \mathrm{OH}(g) \longrightarrow \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \quad \Delta H=+90.7 \mathrm{~kJ}\) (a) Is heat absorbed or released in the course of this reaction? (b) Calculate the amount of heat transferred when 45.0 \(\mathrm{g}\) of \(\mathrm{CH}_{3} \mathrm{OH}(g)\) is decomposed by this reaction at constant pressure. (c) For a given sample of \(\mathrm{CH}_{3} \mathrm{OH}\), the enthalpy change on reaction is \(25.8 \mathrm{~kJ}\). How many grams of hydrogen gas are produced? What is the value of \(\Delta H\) for the reverse of the previous reaction? (d) How many kilojoules of heat are released when \(50.9 \mathrm{~g}\) of \(\mathrm{CO}(\mathrm{g})\) reacts completely with \(\mathrm{H}_{2}(\mathrm{~g})\) to form \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) at constant pressure?

Short answer

(a) Heat is absorbed in the course of this reaction since the enthalpy change is positive (\(\Delta H = +90.7\ \mathrm{kJ}\)). (b) Heat transferred when \(45.0\mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{OH(g)}\) is decomposed is \(127.3\ \mathrm{kJ}\). (c) In this case, \(1.148\ \mathrm{g}\) of hydrogen gas is produced, and the enthalpy change for the reverse reaction is \(\Delta H_{reverse} = -90.7\ \mathrm{kJ}\). (d) Heat released when \(50.9\ \mathrm{g}\) of \(\mathrm{CO(g)}\) reacts completely with \(\mathrm{H}_{2}(g)\) to form \(\mathrm{CH}_{3} \mathrm{OH(g)}\) at constant pressure is \(164.7\ \mathrm{kJ}\).

Step-by-step solution

(a) Heat absorbed or released

For the given reaction, the enthalpy change \(\Delta H = +90.7 \mathrm{~kJ}\). Since the value of \(\Delta H\) is positive, heat is absorbed in the course of this reaction.

(b) Calculating heat transferred

In order to calculate the heat transferred when \(45.0\mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{OH(g)}\) is decomposed, we first need to calculate the number of moles. To find the number of moles, let's divide the mass of the reactant by its molar mass (\(\mathrm{CH}_{3} \mathrm{OH}\): \(\mathrm{C = 12.01, H = 1.01, O = 16.00}\)): Moles of \(\mathrm{CH}_{3} \mathrm{OH} = \frac{45.0\mathrm{~g}}{12.01+4 \times 1.01+16.00} = \frac{45.0\mathrm{~g}}{32.05\mathrm{~g/mol}} = 1.403\mathrm{~mol}\) Since the stoichiometry is \(1:1\) between \(\mathrm{CH}_{3} \mathrm{OH}\) and the reaction, we can use the enthalpy change for the entire reaction, which is \(+90.7\mathrm{~kJ}\) per mole of reactant. Heat transferred, \(q = 1.403\mathrm{~mol}\times 90.7\mathrm{~kJ/mol} = 127.3\mathrm{~kJ}\)

(c) Determine grams of hydrogen gas produced and enthalpy for reverse reaction

We are given the enthalpy change for the reaction (\(\Delta H = +25.8\mathrm{~kJ}\)), and we are asked to find the grams of hydrogen gas produced. From the stoichiometry, we know that the amount of hydrogen gas produced per mole of reaction is the ratio of the moles of hydrogen gas produced to the moles of \(\mathrm{CH}_{3} \mathrm{OH}\) reacted. Since the given enthalpy change is for an unspecified amount of \(\mathrm{CH}_{3} \mathrm{OH}\), let's assign a variable \(x\) to the moles of \(\mathrm{CH}_{3} \mathrm{OH}\), and calculate the moles of \(\mathrm{H}_{2}\) produced. The enthalpy change for this reaction involving \(x\) moles of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(90.7x\) kJ. If it's given that the enthalpy change on reaction is \(25.8\mathrm{~kJ}\), we have: \(90.7x = 25.8\mathrm{~kJ}\\ x = \frac{25.8}{90.7} = 0.284\ \mathrm{mol\ (CH_3OH)}\) Now we can calculate the moles of hydrogen gas produced: Moles of \(\mathrm{H}_{2} = 2 \times 0.284 = 0.568\ \mathrm{mol\ (H_2)}\) Next, we convert the moles of \(\mathrm{H}_{2}\) to grams: Grams of \(\mathrm{H}_{2} = 0.568\mathrm{~mol}\times 2.02\mathrm{~g/mol} = 1.148\mathrm{~g}\) So, \(1.148\mathrm{~g}\) of hydrogen gas is produced. For the reverse reaction, we have: \(\mathrm{CO(g) + 2H_2(g) \to CH_3OH(g)}\) The enthalpy change for the reverse reaction is the negative of the given reaction: \(\Delta H_{reverse} = -\Delta H = -90.7\mathrm{~kJ}\)

(d) Heat released when CO(g) reacts with H2(g)

We are given the amount of \(\mathrm{CO(g)}\) reacting, which is \(50.9\mathrm{~g}\). First, we need to find the moles of \(\mathrm{CO(g)}\): Moles of \(\mathrm{CO(g)} = \frac{50.9\mathrm{~g}}{12.01+16.00} = \frac{50.9\mathrm{~g}}{28.01\mathrm{~g/mol}} = 1.818\mathrm{~mol}\) Since the stoichiometry is \(1:1\) between \(\mathrm{CO}\) and the reverse reaction, we can use the enthalpy change for the reverse reaction, which is \(-90.7\mathrm{~kJ}\) per mole of reactant. Heat released, \(q_{release} = 1.818\mathrm{~mol}\times (-90.7\mathrm{~kJ/mol}) = -164.7\mathrm{~kJ}\) Therefore, \(164.7\mathrm{~kJ}\) of heat is released when \(50.9\mathrm{~g}\) of \(\mathrm{CO(g)}\) reacts completely with \(\mathrm{H_{2}(g)}\) to form \(\mathrm{CH}_{3} \mathrm{OH(g)}\) at constant pressure.

Key concepts

Heat Transfer in Reactions

Understanding heat transfer in chemical reactions is essential to grasping the conservation of energy during these processes. In chemistry, the term enthalpy change, denoted as \(\Delta H\), is used to describe the heat absorbed or released by a system at constant pressure. If \(\Delta H\) is positive, the reaction is endothermic, meaning heat is absorbed from the surroundings. Conversely, when \(\Delta H\) is negative, the reaction is exothermic, releasing heat to the surroundings.

The reaction \(\text{CH}_3 \text{OH}(g) \longrightarrow \text{CO}(g) + 2 \text{H}_2(g)\) with a \(\Delta H=+90.7 \text{kJ}\) is an endothermic process. This positive enthalpy change tells us that the decomposition of methanol into carbon monoxide and hydrogen gas requires the absorption of heat, indicating an input of energy is necessary for the reaction to proceed.

Stoichiometry Calculations

Stoichiometry is the branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. It allows us to predict the amounts of substances consumed and produced by using balanced chemical equations. To perform stoichiometric calculations, it's vital to understand the molar ratio of reactants to products, as it serves as the conversion factor between moles. For instance, in the decomposition of methanol \(\text{CH}_3 \text{OH}\), the balanced chemical equation suggests a direct one-to-one relationship; meaning, for every mole of methanol decomposed, one mole of carbon monoxide and two moles of hydrogen gas are produced.

When given the enthalpy change for a reaction and the mass of a reactant or product, stoichiometry enables us to calculate the heat transferred during the reaction. For example, when we know the enthalpy change for the decomposition of 45 grams of methanol, stoichiometric calculations help us determine the total heat absorbed. These calculations are an integral part of predicting reaction outcomes — essential in both academic contexts and industrial applications.

Molar Mass

Molar mass is a fundamental concept in chemistry, representing the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in a molecule. Knowing the molar mass allows for the conversion of mass to moles, crucial for quantifying the amount of a substance. For example, the molar mass of methanol \(\text{CH}_3 \text{OH}\) is calculated by adding the atomic masses of carbon (12.01 g/mol), oxygen (16.00 g/mol), and four hydrogen atoms (1.01 g/mol each), resulting in a molar mass of 32.04 g/mol.

Utilizing the molar mass is vital when addressing questions related to enthalpy changes in reactions because it bridges the gap between the microscopic scale (molecules and atoms) and the macroscopic scale (grams and kilograms). Whether calculating the number of moles of a substance reacted or determining the amount of heat transfer, molar mass is a key player in comprehending and applying stoichiometric relationships.