Question

Given the reaction:2HgO(s) → 2Hg(l) + O2(g)Calculate the standard enthalpy change ($\Delta H$) and standard entropy change ($\Delta S$) at 298 K, if the standard enthalpy of formation of HgO(s), Hg(l), and O2(g) are -90.8 kJ/mol, 0 kJ/mol, and 0 kJ/mol, respectively.

Answer 1

To calculate the standard enthalpy change  $\Delta H$  for the reaction, we can use the following equation:$\Delta H = \sum \Delta H_f^\circ products  - \sum \Delta H_f^\circ reactants $where $\Delta H_f^\circ$ is the standard enthalpy of formation.For the given reaction:2HgO s   2Hg l  + O2 g The standard enthalpy of formation for the products are:Hg l : 0 kJ/molO2 g : 0 kJ/molThe standard enthalpy of formation for the reactants are:HgO s : -90.8 kJ/molNow, we can plug these values into the equation:$\Delta H = [ 2 \times 0  +  1 \times 0 ] - [ 2 \times -90.8 ]$$\Delta H = 0 -  -181.6 $$\Delta H = 181.6$ kJ/molNow, to calculate the standard entropy change  $\Delta S$ , we need the standard molar entropies  $S^\circ$  of the reactants and products. Unfortunately, these values are not provided in the problem statement. If you can provide the standard molar entropies for HgO s , Hg l , and O2 g , we can calculate the standard entropy change for the reaction.