The standard enthalpy change ΔH o rxn for the thermal decomposition of silver nitrate according to the following equation is 78.67 kJ: AgNO3(s) → AgNO2(s) + 1 2 O2(g) The standard enthalpy of formation of AgNO3(s) is −123.02 kJ/mol. Calculate the standard enthalpy of formation of AgNO2(s).

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]

The equation used to calculate enthalpy change is of a reaction is:

[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]

For the given chemical reaction:

[tex]AgNO_3(s)\rightarrow AgNO_2(s)+\frac{1}{2}O_2(g)[/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(AgNO_2)})+(\frac{1}{2}\times \Delta H^o_f_{(O_2)})]-[(1\times \Delta H^o_f_{(AgNO_3)})][/tex]

We are given:

[tex]\Delta H^o_f_{(AgNO_3)}=-123.02kJ/mol\\\Delta H^o_f_{(O_2)}=0kJ/mol\\\Delta H^o_{rxn}=78.67kJ[/tex]

Putting values in above equation, we get:

[tex]78.67=[(1\times \Delta H^o_f_{(AgNO_2)})+(\frac{1}{2}\times 0)]-[1\times (-123.0))]\\\\\Delta H^o_f_{(AgNO_2)}=-44.35kJ/mol[/tex]

Hence, the enthalpy of the formation of [tex]AgNO_2(s)[/tex] is coming out to be -44.35 kJ/mol.