Some "beetles" defend themselves by spraying hot quinone, C₆H₄O₂(l) , at their enemies. Calculate △H° for this reaction. C₆H₄(OH) ₂(l) + H₂O₂(l) → C₆H₄O₂(l) + 2H₂O(l)
Given: C₆H₄(OH) ₂(l) → C₆H₄O₂(l) + H₂(g) +177.4 kJ
The standard enthalpies of formation of H₂O₂(l) and H₂O(l) are -187.4 and -285.8 kJ/mol, respectively.

Question

Quizplus · Accepted Answer

First, we need to balance the given reaction: 2C₆H₄(OH)₂(l) + 2H₂O₂(l) → 2C₆H₄O₂(l) + 4H₂O(l) Now, we can use Hess's Law to find the enthalpy change of the reaction: △H° = (∑△H°f, products) - (∑△H°f, reactants) △H° = [2(-187.4 kJ/mol) + 4(-285.8 kJ/mol)] - [2(-177.4 kJ/mol)] △H° = -413.6 kJ/mol + 354.8 kJ/mol △H° = -58.8 kJ/mol However, the reaction given in the question is for the formation of H₂(g), while we need the enthalpy change for the formation of H₂O(l). To convert the reaction to the desired one, we can use the enthalpy change of vaporization of water (40.7 kJ/mol) and the enthalpy change of formation of water (-285.8 kJ/mol): H₂(g) + H2O(l) → H₂O(l) + H2(g) △H° = 285.8 kJ/mol - 40.7 kJ/mol = 245.1 kJ/mol Now, we can add the enthalpy change of the above reaction to the enthalpy change of the reaction given in the question to obtain the enthalpy change of the desired reaction: △H° = -58.8 kJ/mol + 245.1 kJ/mol = -206.8 kJ/mol Therefore, the answer is A, and the explanation is as above.

Some "Beetles" Defend Themselves by Spraying Hot Quinone, C6H4O2(l), at Their

Some "Beetles" Defend Themselves by Spraying Hot Quinone, C₆H₄O₂(l), at Their