Find the molar enthalpy of formation for paraffin wax (C2H526)) given the following reaction C2H526) + 38 026) ► → 25 CO26) + 26 H, O)
A H = -14 800 kJ

The molar enthalpy of formation for paraffin wax  is  -2460.5 kJ

Explanation:

The balanced chemical reaction is,

The expression for enthalpy change is,

where,

n = number of moles

(as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

Therefore, the molar enthalpy of formation for paraffin wax  is -2460.5 kJ

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.