At constant pressure and 25°C,what is Δ_rH° for the following reaction 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + H₂O(l)
If the complete consumption of 22.7 g of C₂H₆ liberates -1178 kJ of heat energy?

Question

Quizplus · Accepted Answer

To calculate Δ_rH°, we need to determine the standard enthalpy of formation for each species involved in the reaction. Using the given balanced equation, we can construct the following table: | Species | ΔH°f (kJ/mol) | |---------|---------------| | CS2 | +89.4 | | O2 | 0 | | CO2 | -393.5 | | H2S | -20.1 | | H2O | -285.8 | Using Hess's law, we can calculate the standard enthalpy of the reaction: ΔH°rxn = ΣnΔH°f(products) - ΣnΔH°f(reactants) ΔH°rxn = [4(-393.5) + (-285.8)] - [2(89.4) + 7(0) + 1(-20.1)] ΔH°rxn = -1568.1 kJ/mol-rxn Since the reaction is at constant pressure, ΔH°rxn = ΔU°rxn. Using the given heat energy liberated and the calculated ΔH°rxn, we can solve for the number of moles of the reaction: -1178 kJ/mol-rxn / -1568.1 kJ/mol-rxn = 0.751 mol-rxn Finally, we can calculate the standard entropy change of the reaction using the equation: ΔS°rxn = ΣnS°(products) - ΣnS°(reactants) ΔS°rxn = [4(214.8) + 1(69.9)] - [2(151.1) + 7(205.0) + 1(205.4)] ΔS°rxn = -414.4 J/K-rxn Putting everything together, we can now calculate Δ_rH°: Δ_rH° = ΔH°rxn - TΔS°rxn Δ_rH° = (-1568.1 kJ/mol-rxn) - (298 K)(-414.4 J/K-rxn) Δ_rH° = -3122.5 kJ/mol-rxn Rounding to the nearest ten, we get Δ_rH° = -3120 kJ/mol-rxn, which corresponds to answer choice A.

At Constant Pressure and 25°C,what Is ΔrH° for the Following

At Constant Pressure and 25°C,what Is Δ_rH° for the Following