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Chapter 6: Problem 91

Consider the reaction $$ 2 \mathrm{ClF}_{3}(g)+2 \mathrm{NH}_{3}(g) \longrightarrow \mathrm{N}_{2}(g)+6 \mathrm{HF}(g)+\mathrm{Cl}_{2}(g)\quad\Delta H^{\circ}=-1196 \mathrm{kJ} $$ Calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{ClF}_{3}(g)\)

Short Answer

Expert verified
The standard enthalpy of formation for ClF3(g) is 267.8 kJ/mol.

Step by step solution

01

Determine the standard enthalpies of formation for all substances except ClF3(g)

We can find the standard enthalpies of formation in the literature or a standard reference book. Here are the values for the substances involved in the reaction: ΔHf°(NH3(g)) = -45.9 kJ/mol ΔHf°(N2(g)) = 0 kJ/mol (standard reference for elements in their standard state) ΔHf°(HF(g)) = -273.3 kJ/mol ΔHf°(Cl2(g)) = 0 kJ/mol (standard reference for elements in their standard state)
02

Substitute the values into the formula and solve for ΔHf°(ClF3(g))

We know the standard enthalpy change for the overall reaction, ΔH°, which is -1196 kJ. Using the formula ΔH° = ΣnΔHf°(products) - ΣmΔHf°(reactants) and substituting the values we found earlier: -1196 kJ = (ΔHf°(N2(g)) + 6ΔHf°(HF(g)) + ΔHf°(Cl2(g))) - (2ΔHf°(ClF3(g)) + 2ΔHf°(NH3(g))) -1196 kJ = (0 + 6(-273.3 kJ/mol) + 0) - (2ΔHf°(ClF3(g)) + 2(-45.9 kJ/mol)) Now, we'll solve for ΔHf°(ClF3(g)): -1196 kJ = (-1639.8 kJ) - (2ΔHf°(ClF3(g)) - 91.8 kJ) ΔHf°(ClF3(g)) = (-1196 kJ + 1639.8 kJ + 91.8 kJ)/2 = 267.8 kJ/mol So, the standard enthalpy of formation for ClF3(g) is 267.8 kJ/mol.

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A serving size of six cookies contains 4 g of fat, 20 of carbohydrates, and 2 g of protein. If walking 1.0 mile consumes 170 kJ of energy, how many miles must you walk to burn off enough calories to eat six cookies? Assume the energy content of fats, carbohydrates, and proteins are 8 kcallg, 4 kcallg, and 4 kcallg, respectively.

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