The oxidation of ammonia is illustrated by the following equation: 4 NH 3 (g) + 5 O 2 (g) → 4 NO (g) + 6 H 2 O (g) Calculate the enthalpy of reaction,ΔH Rxn , based on the given standard heats of formation.

Here we're told the oxidation of ammonia is illustrated by the following equation. Here we need to calculate the enthalpy of reaction based on the given standard heats of formation. So we have the heats of formation of each of the compounds. Now it's important to make sure you get the right state as well of the compound within the balanced equation. Notice here I give us water in both its gaseous phase and liquid phase. So make sure you go back and take a look at the equation. In the equation, we're dealing with gaseous water, which means we shouldn't even look at this liquid form of water. That's the wrong one to use. So always be, cognizant of that. Right. So here we're going to figure out what our delta h of reaction is. So here, remember it's products minus reactants. So what we're gonna do here is we're gonna look at the products first. We have 4 moles of n o, Each n o gas has an enthalpy of formation equal to 90.3 kilojoules per mole, Plus the other product, which is 6 moles of h two gas, not liquid, and it is negative 241.8 kilojoules per mole. Minus your reactants, So we have 4 moles of n h 3, each one is negative 45.9 kilojoules per mole. Now oxygen is in its natural or standard state, when you're in your natural standard state your enthalpy of formation is equal to 0. We don't even need to include it within our formula because anything times 0 is 0. We see here that our moles will cancel out, and we'll be left with kilojoules. Our products when we multiply and add them together gives us negative 1089.6 kilojoules minus your reactants which comes out to negative 183.6 kilojoules. So minus on minus is really adding, so when you you do so it's really negative 1089.6 kilojoules plus 183 0.6 kilojoules, and that gives me negative 906 kilojoules as my enthalpy of reaction. So remember, just utilize the formula, take a look at the correct form to find the enthalpies of formation, and you can find Delta H of reaction whether it's under standard conditions or non standard conditions in this

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8. Thermochemistry

Enthalpy of Formation

General Chemistry

8. Thermochemistry

Enthalpy of Formation

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Multiple Choice

The oxidation of ammonia is illustrated by the following equation:
4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g)
Calculate the enthalpy of reaction,ΔH_Rxn, based on the given standard heats of formation.

-906 kJ

-1273.2 kJ

1089.6 kJ

-183.6 kJ

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Verified step by step guidance

Identify the balanced chemical equation for the reaction: 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g).

Use the standard heats of formation (ΔHf°) from the table: NH3 (g) = -45.9 kJ/mol, O2 (g) = 0.0 kJ/mol, NO (g) = 90.3 kJ/mol, H2O (g) = -241.8 kJ/mol.

Apply the formula for the enthalpy change of the reaction: ΔHRxn = Σ(ΔHf° of products) - Σ(ΔHf° of reactants).

Calculate the total ΔHf° for the products: (4 mol × 90.3 kJ/mol for NO) + (6 mol × -241.8 kJ/mol for H2O).

Calculate the total ΔHf° for the reactants: (4 mol × -45.9 kJ/mol for NH3) + (5 mol × 0.0 kJ/mol for O2) and subtract this from the total ΔHf° of the products to find ΔHRxn.