What is the enthalpy of reaction D H rxn for the following reaction?
SiO₂ (s) + 4 HF (g) SiF₄ (g) + 2 H₂ O ( ) D H rxn = ??
D H f (SiO₂ ) = - 910.9 kJ/mol D H f (HF) = - 273.0 kJ/mol D H f (SiF₄ ) = - 1614.9 kJ/mol D H f (H₂ O) = - 285.8 kJ/mol

Question

What is the enthalpy of reaction D H rxn for the following reaction?
SiO₂ (s) + 4 HF (g) SiF₄ (g) + 2 H₂ O (

What is the enthalpy of reaction D H rxn for the following reaction? SiO₂ (s) + 4 HF (g) SiF<sub>4</sub> (g) + 2 H₂ O ( ) D H rxn = ?? D H f (SiO₂ ) = - 910.9 kJ/mol D H f (HF) = - 273.0 kJ/mol D H f (SiF<sub>4</sub> ) = - 1614.9 kJ/mol D H f (H₂ O) = - 285.8 kJ/mol A) - 183.6 kJ/mol B) - 716.8 kJ/mol C) - 1002.6 kJ/mol D) - 3084.6 kJ/mol E) - 4189.4 kJ/mol

) D H rxn = ??
D H f (SiO₂ ) = - 910.9 kJ/mol D H f (HF) = - 273.0 kJ/mol D H f (SiF₄ ) = - 1614.9 kJ/mol D H f (H₂ O) = - 285.8 kJ/mol

Quizplus · Accepted Answer

To find the enthalpy of reaction, we need to determine the sum of the enthalpies of the products minus the sum of the enthalpies of the reactants, each multiplied by their respective stoichiometric coefficients.

D H rxn = [n(D H f products)] - [n(D H f reactants)]
D H rxn = [(1)(-1614.9 kJ/mol) + 2(-285.8 kJ/mol)] - [(1)(-910.9 kJ/mol) + 4(-273.0 kJ/mol)]
D H rxn = - 183.6 kJ/mol

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