Calculate AG (13/mol) for the following reaction (at 298 K): 2 Hg (g) +0,(9) + 2 HgO(s)
(AG° = -180.8 kJ/mol), if the partial pressure of Hg and o are 0.025 and 0.037 atm,
respectively.
(R = 0.082 atm. L/K. mol = 8.314 J/K. mol)
(2.5 Points)​

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

i think its 100.

hope this !