Question

A.Using standard heats of formation, calculate the standard enthalpy change for the following reaction.

N₂(g) + 3H₂(g) = 2NH₃(g)

B.Using standard heats of formation, calculate the standard enthalpy change for the following reaction.

CaCO₃(s) = CaO(s) + CO₂(g)

C. A scientist measures the standard enthalpy change for the following reaction to be -2910.6 kJ:

2C₂H₆(g) + 7 O₂(g) = 4CO₂(g) + 6 H₂O(g)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H₂O(g) is ____ kJ/mol.

D. A scientist measures the standard enthalpy change for the following reaction to be -1051.8 kJ :

2H₂S(g) + 3 O₂(g) = 2H₂O(g) + 2 SO₂(g)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H₂O(g) is _____ kJ/mol.

Answer 1

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Answer 2

To calculate the standard enthalpy change for a reaction using standard heats of formation, we use the following equation:

ΔH° = Σ(n * ΔH°f(products)) - Σ(m * ΔH°f(reactants))

Where: ΔH° is the standard enthalpy change for the reaction. n and m are the stoichiometric coefficients of the products and reactants, respectively. ΔH°f is the standard heat of formation for each substance.

A. For the reaction: N2(g) + 3H2(g) → 2NH3(g)

The standard heats of formation are as follows: ΔH°f(N2) = 0 kJ/mol ΔH°f(H2) = 0 kJ/mol ΔH°f(NH3) = -46.11 kJ/mol

Using the equation, we get: ΔH° = (2 * ΔH°f(NH3)) - (ΔH°f(N2) + 3 * ΔH°f(H2)) ΔH° = (2 * -46.11 kJ/mol) - (0 kJ/mol + 3 * 0 kJ/mol) ΔH° = -92.22 kJ/mol

B. For the reaction: CaCO3(s) → CaO(s) + CO2(g)

The standard heats of formation are as follows: ΔH°f(CaCO3) = -1207.7 kJ/mol ΔH°f(CaO) = -635.1 kJ/mol ΔH°f(CO2) = -393.5 kJ/mol

Using the equation, we get: ΔH° = (ΔH°f(CaO) + ΔH°f(CO2)) - ΔH°f(CaCO3) ΔH° = (-635.1 kJ/mol + -393.5 kJ/mol) - (-1207.7 kJ/mol) ΔH° = -27.1 kJ/mol

C. For the reaction: 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

Given ΔH° = -2910.6 kJ/mol

The standard heats of formation for CO2 and C2H6 are as follows: ΔH°f(CO2) = -393.5 kJ/mol ΔH°f(C2H6) = -84.68 kJ/mol

Using the equation, we can rearrange it to solve for ΔH°f(H2O): ΔH°f(H2O) = (1/6) * [4 * ΔH°f(CO2) - 2 * ΔH°f(C2H6) + ΔH°]

ΔH°f(H2O) = (1/6) * [4 * (-393.5 kJ/mol) - 2 * (-84.68 kJ/mol) - 2910.6 kJ/mol] ΔH°f(H2O) = (1/6) * [-1574 kJ/mol + 169.36 kJ/mol - 2910.6 kJ/mol] ΔH°f(H2O) = (1/6) * [-3315.24 kJ/mol] ΔH°f(H2O) ≈ -552.54 kJ/mol

D. For the reaction: 2H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g)

Given ΔH° = -1051.8 kJ/mol

The standard heats of formation for H2O and H2S are as follows: ΔH°f(H2O) = -241.82 kJ/mol ΔH°f(H2S) = -20.63 kJ/mol

Using the equation, we can rearrange it to solve for ΔH°f(H2O): ΔH°f(H2O) = (1/2) * [2 * ΔH°f(H2S) - 2 * ΔH° + ΔH°f(SO2)]

ΔH°f(H2O) = (1/2) * [2 * (-20.63 kJ/mol) - 2 * (-1051.8 kJ/mol) + ΔH°f(SO2)] ΔH°f(H2O) = (1/2) * [-41.26 kJ/mol + 2103.6 kJ/mol + ΔH°f(SO2)] ΔH°f(H2O) = (1/2) * [2062.34 kJ/mol + ΔH°f(SO2)] ΔH°f(H2O) ≈ 1031.17 kJ/mol

The standard enthalpy of formation of H2O(g) is approximately 1031.17 kJ/mol.