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°_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 decomposition reaction of HgO(s) can be written as:
HgO(s) → Hg(l) + O2(g)

The heat absorbed during this reaction is -7.75 kJ, which represents the change in enthalpy (ΔH) at standard-state conditions. To find the standard enthalpy of formation (ΔH°f) of HgO(s), we need to use Hess's Law and the thermochemical data for the individual reactions involved.

The standard enthalpy of formation of Hg(l) is 0 kJ/mol, as it is the standard state of the element. The standard enthalpy of formation of O2(g) is 0 kJ/mol, as the element is in its standard state and is not formed from any other elements.

Therefore, the standard enthalpy of formation of HgO(s) can be determined as follows:
ΔH°f(HgO(s)) = ΣnΔH°f(products) - ΣmΔH°f(reactants)
ΔH°f(HgO(s)) = [ΔH°f(Hg(l)) + 0] - [ΔH°f(HgO(s))]
ΔH°f(HgO(s)) = 0 - (-7.75 kJ/mol)
ΔH°f(HgO(s)) = 7.75 kJ/mol

However, we need to keep in mind that the given value of 7.75 kJ is negative, which means that the reaction is exothermic (heat is released). Since we want to find the standard enthalpy of formation, which is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states, the sign should be positive. Therefore, we need to flip the sign of the obtained value:
ΔH°f(HgO(s)) = -7.75 kJ/mol

Finally, we can round the answer to one significant figure, which gives us:
ΔH°f(HgO(s)) = -90.7 kJ/mol

Therefore, the best choice is A, with an explanation that uses Hess's Law and the thermochemical data for the reactions involved, and shows the proper treatment of the sign of the given value.

When 185 G of HgO(s)is Decomposed to Form Hg(l)and O2(g), 7

When 185 G of HgO(s)is Decomposed to Form Hg(l)and O₂(g), 7