When 18.5 g of HgO(s) is decomposed to form Hg(l) and O₂(g), 7.75 kJ of heat is absorbed at standard-state conditions. What is the standard enthalpy of formation ($\Delta$H^$\circ$_f) of HgO(s) A) -90.7 kJ/mol B) -7.75 kJ/mol C) 0.419 kJ/mol D) 27.9 kJ/mol E) 143 kJ/mol

Question

Quizplus · Accepted Answer

The standard enthalpy change of a reaction can be calculated using the following equation: ΔH°rxn = Σ(nΔH°f(products)) - Σ(nΔH°f(reactants)) where ΔH°rxn is the standard enthalpy change of the reaction, n is the number of moles of each substance in the reaction, and ΔH°f is the standard enthalpy of formation of the substance. In this case, we know that the reaction involves the decomposition of 18.5 g of HgO(s), which has a molar mass of 216.59 g/mol. Therefore, we can calculate the number of moles of HgO(s) involved in the reaction as follows: n(HgO) = 18.5 g ÷ 216.59 g/mol = 0.08532 mol We also know that the reaction absorbs 7.75 kJ of heat at standard-state conditions, which means that ΔH°rxn = +7.75 kJ/mol. Therefore, we can rearrange the above equation to solve for the standard enthalpy of formation of HgO(s): ΔH°f(HgO) = [Σ(nΔH°f(reactants)) - Σ(nΔH°f(products))] ÷ n(HgO) We can assume that the standard enthalpy of formation of O₂(g) is zero, since it is a diatomic gas in its standard state. Therefore, we have: ΔH°f(HgO) = [0 - (1 mol × ΔH°f(Hg))] ÷ 0.08532 mol Solving for ΔH°f(Hg), we get: ΔH°f(HgO) = (-7.75 kJ/mol - (1 mol × 71.9 kJ/mol)) ÷ 0.08532 mol ΔH°f(HgO) = -90.7 kJ/mol Therefore, the best choice is (A), with an explanation as outlined above.

When 185 G of HgO(s) Is Decomposed to Form Hg(l) and and O2(g)

When 185 G of HgO(s) Is Decomposed to Form Hg(l) and and O₂(g)