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

Question

Quizplus · Accepted Answer

The reaction given is a combination of two reactions. The first reaction is provided with an enthalpy change of +177.4 kJ/mol, and it involves the formation of CS16H14O(l) and H2O(g) from CS16H14(OH)2(l). The second part of the reaction involves the formation of 2 H2O(l) from 2 H2O(g), which can be calculated using the enthalpy of formation of H2O(l). The enthalpy change for the reaction of interest can be calculated by subtracting the enthalpy of formation of the products from the reactants. Given the enthalpies of formation for H2O(l) and H2SO(l), the calculation involves adjusting for the phase change of water and accounting for the given reaction enthalpy. The correct calculation leads to an enthalpy change of -206.8 kJ/mol.

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

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