Answer:
The enthalpy change for the given reaction is 424 kJ.
Explanation:
[tex]2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?[/tex]
We have :
Enthalpy changes of formation of following s:
[tex]\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol[/tex]
[tex]\Delta H_{f,CuO}=-157.3 kJ/mol[/tex]
[tex]\Delta H_{f,NO_2}= 33.2 kJ/mol[/tex]
[tex]\Delta H_{f,O_2}= 0 kJ/mol[/tex] (standard state)
[tex]\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)][/tex]
The equation for the enthalpy change of the given reaction is:
[tex]\Delta H_{rxn}[/tex] =
[tex]=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})[/tex]
[tex]\Delta H_{rxn}[/tex]=
[tex](2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)[/tex]
[tex]\Delta H_{rxn}=424 kJ[/tex]
The enthalpy change for the given reaction is 424 kJ.
