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Find the ΔHr° for the reaction of three moles of potassium hydroxide and one mole of phosphoric acid that produces one mole of potassium phosphate and three moles of water.

3KOH(aq)+H₃PO₄(aq) ⟶ K₃PO₄(aq)+3H₂O(l)

A. -2726. 11 kJ
B. -81. 4 kJ
C. -55. 34 kJ
D. -2807. 49 kJ
E. -5533. 6 kJ

User Magqq
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1 Answer

7 votes

Final answer:

Without the standard enthalpies of formation for the reactants and products, the standard enthalpy of reaction (ΔHr°) for the given reaction cannot be determined from the provided information. Therefore, we cannot provide the correct option for the enthalpy change of the reaction between potassium hydroxide and phosphoric acid.

Step-by-step explanation:

The student is seeking the standard enthalpy of reaction (ΔHr°) for the reaction between potassium hydroxide and phosphoric acid to produce potassium phosphate and water. To calculate the enthalpy change of this reaction, we generally use the standard enthalpy of formation values for all the reactants and products involved. However, the information provided does not include these values, and thus we cannot calculate ΔHr° directly from the given information.

It is important to note that without the standard enthalpies of formation, the value for ΔHr° cannot be determined from the provided text.

The data regarding ionization constants of polyprotic acids, titration of phosphoric acid with potassium hydroxide, and other examples of enthalpy changes are not directly applicable to calculating the enthalpy change of the given reaction. Therefore, we are currently unable to provide the correct option from A to E for the enthalpy change of the reaction 3KOH(aq) + H₃PO₄(aq) → K₃PO₄(aq) + 3H₂O(l).

User Anthoni
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