From the enthalpies of reaction 2 C(s) + O 2 (g) → 2 CO(g) ΔH = -221.0 kJ 2 C(s) + O 2 (g) + 4 H 2 (g) → 2 CH 3 OH(g) ΔH = -402.4 kJ Calculate ΔH for the reaction CO(g) + 2 H 2 (g) → CH 3 OH(g)

hey everyone in this example, we're given the below data and we need to find our entropy change for our final reaction here. So our first step is to recognize how our first given reaction compares to our final reaction and we should recognize that our two moles of nitrogen gas should be on the react inside of our final equation. So we're going to want to start by saying that we should reverse this reaction and that includes reversing our sign for entropy. We also should recognize that we have too many moles of all of our agents. And so we're going to go ahead and divide this entire equation by two so we can go ahead and just multiply this by one half. So what that would give us for our manipulated equation is now we have one mole of N two gas on the reactant side Added to three moles of water on our react inside as well. Then this is going to produce 3/2 moles of our oxygen gas And now two moles of our ammonia gas. We also want to go ahead and recalculate our entropy value here. And so we would take our negative 1530 kg jewels And multiply that by 1/2 And that's going to give us a value of 765 kg jewels when we reverse the sign of our entropy, which should make it positive here. So this is our first manipulated equation and now we want to go ahead and analyze our second equation. So in comparison to the final reaction, our second equation has all of the re agents on the proper sides of our equation. However, we have too many moles of all of our agents. So we're going to take these re agents and we're going to go ahead and Multiply everything by 1/2. So that would also include multiplying our entropy value by 1/2. And so now we would have a new equation where we have 5/2 moles of our oxygen gas Added to two moles of our ammonia gas, producing two moles of our nitrogen monoxide gas. And now we have three moles of our liquid water. And now our entropy value here is going to be negative 1170 kg joules times one half. And that's going to give us negative 5 85 kg joules. So we've maintained our sign here for entropy due to the fact that we haven't reversed our equation. We just we're multiplying it by one half. So that just gave us negative 5 85 kg joules. And now our next step is to go ahead and add these two equations together to see how they give us our final entropy value. And if they match up to our final equation, so we should recognize that we can cancel out our three moles of water because we have them in both equations. We also can go ahead and get rid of the two moles of our Ammonia gas in both equations. And so this is going to give us our final equation where we have one mole of co two gas Added to one mole of nitrogen gas, producing two moles of nitrogen monoxide gas. And when we take the two entropy values here and add them up together, this is positive 7 65 plus negative 5 85 kg joules, Which is going to give us a final entropy value of 180 kg jewels. And so this is going to be our final answer as our change in entropy for our final given reaction here. So I hope that everything we went through was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

Ch.5 - Thermochemistry

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Brown 15th Edition

Ch.5 - Thermochemistry

Problem 64

Chapter 5, Problem 64

From the enthalpies of reaction 2 C(s) + O₂(g) → 2 CO(g) ΔH = -221.0 kJ 2 C(s) + O₂(g) + 4 H₂(g) → 2 CH₃OH(g) ΔH = -402.4 kJ Calculate ΔH for the reaction CO(g) + 2 H₂(g) → CH₃OH(g)

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Identify the target reaction: CO(g) + 2 H_2(g) → CH_3OH(g).

Use Hess's Law, which states that the total enthalpy change for a reaction is the same, no matter how many steps the reaction is carried out in.

Write the given reactions and their enthalpy changes: 1) 2 C(s) + O_2(g) → 2 CO(g), ΔH = -221.0 kJ; 2) 2 C(s) + O_2(g) + 4 H_2(g) → 2 CH_3OH(g), ΔH = -402.4 kJ.

Manipulate the given reactions to derive the target reaction. Divide both given reactions by 2 to match the stoichiometry of the target reaction.

Subtract the enthalpy of the first reaction from the second to find the enthalpy change for the target reaction.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Enthalpy of Reaction

Enthalpy of reaction (ΔH) is the heat change associated with a chemical reaction at constant pressure. It indicates whether a reaction is exothermic (releases heat, ΔH < 0) or endothermic (absorbs heat, ΔH > 0). Understanding ΔH is crucial for predicting the energy changes during reactions and for applying Hess's law to calculate unknown enthalpies.

Hess's Law

Hess's Law states that the total enthalpy change for a reaction is the sum of the enthalpy changes for individual steps, regardless of the pathway taken. This principle allows chemists to calculate the enthalpy change for a reaction that may be difficult to measure directly by using known enthalpy changes of related reactions.

Stoichiometry in Reactions

Stoichiometry involves the quantitative relationships between the reactants and products in a chemical reaction. It is essential for balancing chemical equations and determining the amounts of substances consumed or produced. In this context, stoichiometry helps in relating the enthalpy changes of the given reactions to find the ΔH for the target reaction.

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Related Practice

Textbook Question

Under constant-volume conditions, the heat of combustion of benzoic acid (C₆H₅COOH) is 26.38 kJ/g. A 2.760-g sample of benzoic acid is burned in a bomb calorimeter. The temperature of the calorimeter increases from 21.60 to 29.93 °C. c. Suppose that in changing samples, a portion of the water in the calorimeter were lost. In what way, if any, would this change the heat capacity of the calorimeter?

Textbook Question

Consider the following hypothetical reactions: A → B ΔH = +30 kJ B → C ΔH = +60 kJ (b) Construct an enthalpy diagram for substances A, B, and C, and show how Hess's law applies.

Textbook Question

Calculate the enthalpy change for the reaction P₄O₆(s) + 2 O₂(g) → P₄O₁₀(s) given the following enthalpies of reaction: P₄(s) + 3 O₂(g) → P₄O₆(s) ΔH = -1640.1 kJ P₄(s) + 5 O₂(g) → P₄O₁₀(s) ΔH = -2940.1 kJ

Textbook Question

The concentration of alcohol 1CH3CH2OH2 in blood, called the 'blood alcohol concentration' or BAC, is given in units of grams of alcohol per 100 mL of blood. The legal definition of intoxication, in many states of the United States, is that the BAC is 0.08 or higher. What is the concentration of alcohol, in terms of molarity, in blood if the BAC is 0.08?

Textbook Question

Given the data N₂(g) + O₂(g) → 2 NO(g) ΔH = +180.7 kJ 2 NO(g) + O₂(g) → 2 NO₂(g) ΔH = -113.1 kJ 2 N₂O(g) → 2 N₂(g) + O₂(g) ΔH = -163.2 kJ use Hess's law to calculate ΔH for the reaction N₂O(g) + NO₂(g) → 3 NO(g)

Textbook Question

We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethane: 2 CH4(g) → C2H6(g) + H2(g) Calculate the ΔH° for this reaction using the following thermochemical data: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ΔH° = -890.3 kJ 2 H2(g) + O2(g) → 2 H2O(l) H° = -571.6 kJ 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l) ΔH° = -3120.8 kJ

From the enthalpies of reaction 2 C(s) + O2(g) → 2 CO(g) ΔH = -221.0 kJ 2 C(s) + O2(g) + 4 H2(g) → 2 CH3OH(g) ΔH = -402.4 kJ Calculate ΔH for the reaction CO(g) + 2 H2(g) → CH3OH(g)

Verified video answer for a similar problem:

Key Concepts

Enthalpy of Reaction

Hess's Law

Stoichiometry in Reactions

From the enthalpies of reaction 2 C(s) + O₂(g) → 2 CO(g) ΔH = -221.0 kJ 2 C(s) + O₂(g) + 4 H₂(g) → 2 CH₃OH(g) ΔH = -402.4 kJ Calculate ΔH for the reaction CO(g) + 2 H₂(g) → CH₃OH(g)