Consider the following equation: 2 ClF 3 (g) + 2 NH 3 (g) → 1 N 2 (g) + 6 HF (g) + 6 Cl 2 (g)ΔH rxn = –1196 kJ Determine the standard enthalpy of formation for chlorine trifluoride, ClF 3 .

So here it says, consider the following reaction. Determine the standard enthalpy of formation for Chlorine Trifluoride. Alright. So they're giving us all the compounds within our balanced equation, and they're giving us already the enthalpy of the reaction. So remember, we're gonna say enthalpy of a reaction equals products minus reactants. So here that's gonna equal, they already give us the enthalpy of reaction, so that's negative 1196 kilojoules equals, for our products, n 2 is in its natural state. Remember when you're in your natural state your enthalpy of formation is equal to 0. We don't need to include it within our calculations because anything times 0 is 0, so we're gonna have 6 moles of h f. Right? And then each mole is negative 273 kilojoules per mole plus our other product, c l 2, is also in its natural state, so its enthalpy of formation is also equal to 0, so we don't need to include it. Okay. Minus our reactants. So our reactants we have are 2 moles of clf3. We don't know what its enthalpy of formation is, that's what we're looking to find, plus 2 moles of n h 3, each one is negative 45.9 kilojoules per mole. Here moles will cancel out and we'll have kilojoules. So this is negative 1196 kilojoules equals negative 1638, when we have 6 times negative 273, minus 2x plus negative 96.8. Then that's gonna be negative 1196 kilojoules equals negative 1638. This negative gets distributed, right? So this is going to be negative 2 x, and negative times a negative is a positive, so plus 96 point 8. We're gonna have negative 1638 and positive 96.8 adding together. When we do that, we're gonna get 91. We're gonna get 91 oops. Actually, what we're gonna do is we don't need to add them together just yet. We can just add 1638 to both sides first. Okay. So we'll do 1638 to both sides here. This is 91.8. So this is gonna be 91.8. So when we do that we're gonna get here 442 equals negative 2x +91.8 plus 91.8. Subtract 91.8 from both sides. So that's going to give me 350.2 equals negative 2 x. Divide both sides by negative 2. So x will equal negative 1750 or 175.1. Since this is an enthalpy of formation its units will be in kilojoules per mole. So this is the enthalpy of formation for each mole of chlorine trichloride, so that will be our final answer.

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8. Thermochemistry

Enthalpy of Formation

General Chemistry

8. Thermochemistry

Enthalpy of Formation

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Multiple Choice

Consider the following equation:
2 ClF₃(g) + 2 NH₃(g) → 1 N₂(g) + 6 HF (g) + 6 Cl₂(g)ΔH_rxn = –1196 kJ
Determine the standard enthalpy of formation for chlorine trifluoride, ClF₃.

-175.1 kJ

350.2 kJ

442.0 kJ

-1638 kJ

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Verified step by step guidance

Identify the given chemical reaction: 2 ClF3(g) + 2 NH3(g) → 1 N2(g) + 6 HF(g) + 6 Cl2(g) with ΔHrxn = –1196 kJ.

Use the formula for the standard enthalpy change of reaction: ΔHrxn = ΣΔHf(products) - ΣΔHf(reactants).

List the standard enthalpies of formation for the products and reactants: ΔHf(NH3) = -45.9 kJ/mol, ΔHf(N2) = 0 kJ/mol, ΔHf(HF) = -273 kJ/mol, ΔHf(Cl2) = 0 kJ/mol.

Calculate the total enthalpy of formation for the products: (1 mol N2 * 0 kJ/mol) + (6 mol HF * -273 kJ/mol) + (6 mol Cl2 * 0 kJ/mol).

Calculate the total enthalpy of formation for the reactants: (2 mol ClF3 * ΔHf(ClF3)) + (2 mol NH3 * -45.9 kJ/mol) and solve for ΔHf(ClF3) using the equation from step 2.