Hydrazine (N 2 H 4 ) is a fuel used by some spacecraft. It is normally oxidized by N 2 O 4 according to the equation: N 2 H 4 (l) + N 2 O 4 (g) → 2 N 2 O (g) + 2 H 2 O (g) Calculate ΔH° rxn for this reaction using standard enthalpies of formation.

Identify the standard enthalpies of formation (ΔH° f ) for each of the reactants and products involved in the reaction. These values are typically found in a chemistry reference or textbook. For this reaction, you need the ΔH° f for N 2 H 4 (l), N 2 O 4 (g), N 2 O (g), and H 2 O (g).. Write down the balanced chemical equation: N 2 H 4 (l) + N 2 O 4 (g) → 2 N 2 O (g) + 2 H 2 O (g).. Apply the formula for calculating the standard enthalpy change of the reaction (ΔH° rxn ): ΔH° rxn = ΣΔH° f (products) - ΣΔH° f (reactants). Remember to multiply the ΔH° f of each substance by its stoichiometric coefficient in the balanced equation.. Substitute the standard enthalpies of formation into the equation. Make sure to use the correct signs (positive or negative) as provided by the standard enthalpies of formation values.. Calculate the sum of the standard enthalpies of formation for the products and reactants separately, then find the difference between these sums to determine ΔH° rxn .

Hydrazine (N 2 H 4 ) is a fuel used by some spacecraft. It is normally oxidized by N 2 O 4 according to the equation: N 2 H 4 (l) + N 2 O 4 (g) → 2 N 2 O (g) + 2 H 2 O (g) Calculate ΔH° rxn for this reaction using standard enthalpies of formation.

Hi everyone for this problem, it says ammonia is the raw material used in the commercial production of nitric acid. The first step of the process is the oxidation of ammonia and were given the chemical equation Using standard entropy of formation, we need to calculate the standard entropy change for this reaction. So let's go ahead and write out what that means. And so the standard entropy change of our reaction is going to equal the sum of our products, minus the sum of our reactant. And I'll explain what that means. And so we're going to need our standard heats of formation for our each thing in our reaction in order to calculate this. So let's go ahead and look up those values because it's going to be necessary. And so the standard heat of formation for our ammonia is negative 46. killer jewels per mole. The standard heat of formation for R. N. O. Is going to equal positive 90.29 killer jules Permal. And the standard heat of formation for H20 gas is going to equal negative 20 sorry, negative 241 .8 Killer Jewels per more. And we need to pay attention to see if we have anything in its elemental state because anything in its elemental state Is going to equal zero. So for this problem we have oxygen and its elemental state. And so that heat of formation is zero. So let's continue with solving for our standard entropy change of our reaction. So we have The some of our products. And so let's take a look at our products, we have no gas and H20 gas. And so we're going to multiply how many moles we're going to take the some of our products minus the sum of our reactant. So we're going to take how many moles we have of each of our product and multiply it by its standard heat of formation. So let's start off with our H 20 gas first. So we have six moles of H 20 gas and we're going to multiply it by its standard heat of formation, which is negative 0.8 killer jewels Permal. And we're going to add because we're taking the some of our products, we have four moles of N. O gas, Multiplied by its standard heat of formation is 90 0.29 Killer jules Permal. So that's the sum of our products minus the sum of our reactant. And so here we said that oxygen is in its elemental state. So we can still write it out but we know it's going to be zero. So we have five moles of oxygen times zero kg joules per mole plus Formals of Ammonia and its standard heat of formation is negative 46.1 killer jewels per mole. Okay, I'm running out of space here and so we can go ahead and continue to solve this. We know that our five moles of oxygen is going to equal zero so we can go ahead and eliminate that now. But we wrote it out just so that we can see and so when we simplify this, we get our standard and they'll be of our reaction is going to equal. When we simplify the some of our products, we get negative 1000 89 point 64 killer jewels per mole minus -184 Point four killer jewels. So that's the sum of our products minus the sum of our reactant. And we get a final standard. And there'll be change of our reaction to be -905. killer jewels per this is our standard entropy change for this reaction. That's the end of this problem. I hope this was helpful.

Ch.7 - Thermochemistry

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Tro 5th Edition

Ch.7 - Thermochemistry

Problem 85

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Chapter 7, Problem 85

Hydrazine (N₂H₄) is a fuel used by some spacecraft. It is normally oxidized by N₂O₄ according to the equation: N₂H₄(l) + N₂O₄ (g) → 2 N₂O (g) + 2 H₂O (g) Calculate ΔH°_rxn for this reaction using standard enthalpies of formation.

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Identify the standard enthalpies of formation (ΔH°f) for each of the reactants and products involved in the reaction. These values are typically found in a chemistry reference or textbook. For this reaction, you need the ΔH°f for N2H4 (l), N2O4 (g), N2O (g), and H2O (g).

Write down the balanced chemical equation: N2H4 (l) + N2O4 (g) → 2 N2O (g) + 2 H2O (g).

Apply the formula for calculating the standard enthalpy change of the reaction (ΔH°rxn): ΔH°rxn = ΣΔH°f(products) - ΣΔH°f(reactants). Remember to multiply the ΔH°f of each substance by its stoichiometric coefficient in the balanced equation.

Substitute the standard enthalpies of formation into the equation. Make sure to use the correct signs (positive or negative) as provided by the standard enthalpies of formation values.

Calculate the sum of the standard enthalpies of formation for the products and reactants separately, then find the difference between these sums to determine ΔH°rxn.

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Key Concepts

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

Standard Enthalpy of Formation

The standard enthalpy of formation (ΔH°f) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. It is a crucial value used in thermodynamics to calculate the overall energy change in chemical reactions. Each substance has a specific ΔH°f value, which can be found in tables, and is essential for determining the enthalpy change of a reaction.

Hess's Law

Hess's Law states that the total enthalpy change for a reaction is the same, regardless of the number of steps taken to achieve the reaction. This principle allows for the calculation of the enthalpy change of a reaction by summing the enthalpy changes of individual steps, making it possible to use standard enthalpies of formation to find ΔH°rxn for complex reactions.

Enthalpy Change of a Reaction (ΔH°rxn)

The enthalpy change of a reaction (ΔH°rxn) is the difference between the total enthalpy of the products and the total enthalpy of the reactants under standard conditions. It indicates whether a reaction is exothermic (releases heat, ΔH°rxn < 0) or endothermic (absorbs heat, ΔH°rxn > 0). Calculating ΔH°rxn using standard enthalpies of formation involves subtracting the sum of the ΔH°f values of the reactants from that of the products.