Hess’s Law How do we cope with energy changes in energy transfers that are difficult to measure using calorimetry?

Hess’s Law! The value of the ∆H (change in enthalpy) for any reaction can be obtained by adding the ∆H values for reaction steps or components The value of the ∆H (change in enthalpy) for any reaction can be obtained by adding the ∆H values for reaction steps or components

Example: N 2 + O 2 2NO + energy N 2 + O 2 2NO + energy This is an exothermic reaction that is not easily done in a calorimeter This is an exothermic reaction that is not easily done in a calorimeter How can we determine the molar ∆H combustion for nitrogen??? How can we determine the molar ∆H combustion for nitrogen???

Solution Method: One method would be to find alternative reactions that would “add” up to the over all reaction we are interested in. One method would be to find alternative reactions that would “add” up to the over all reaction we are interested in. (Recall doing this in math when you wanted to solve for the intercept of two equations!) (Recall doing this in math when you wanted to solve for the intercept of two equations!)

Component Reactions Example: ½ N 2(g) + O 2(g) NO 2(g) ∆H = +34kJ ½ N 2(g) + O 2(g) NO 2(g) ∆H = +34kJ NO (g) + ½ O 2(g) NO 2(g) ∆H = -56kJ NO (g) + ½ O 2(g) NO 2(g) ∆H = -56kJ These two known and measurable reactions can be re-arranged to “add up” to the original reaction we are interested in! These two known and measurable reactions can be re-arranged to “add up” to the original reaction we are interested in!

Solution N 2 + O 2 2NO Target reaction N 2 + O 2 2NO Target reaction ½ N 2(g) + O 2(g) NO 2(g) ½ N 2(g) + O 2(g) NO 2(g) Note that we want nitrogen as a reactant!Note that we want nitrogen as a reactant! This equation is fine in this order, only requires multiplying x2 (gets rid of ½)This equation is fine in this order, only requires multiplying x2 (gets rid of ½) NO (g) + ½ O 2(g) NO 2(g) NO (g) + ½ O 2(g) NO 2(g) Note that we want nitrogen oxide as a product!Note that we want nitrogen oxide as a product! This equation needs to be “flipped” and multiplied x2 (need 2 NO as product)This equation needs to be “flipped” and multiplied x2 (need 2 NO as product)

Result… N 2(g) + 2 O 2(g) 2 NO 2(g) ∆H = +68kJ N 2(g) + 2 O 2(g) 2 NO 2(g) ∆H = +68kJ 2NO 2(g) 2 NO (g) + O 2(g) ∆H = +112kJ 2NO 2(g) 2 NO (g) + O 2(g) ∆H = +112kJ We can cancel out the products that are reactants in the second equation! We can cancel out the products that are reactants in the second equation!

cntd N 2(g) + 2 O 2(g) 2 NO 2(g) ∆H = +68kJ N 2(g) + 2 O 2(g) 2 NO 2(g) ∆H = +68kJ 2NO 2(g) 2 NO (g) + O 2(g) ∆H = +112kJ 2NO 2(g) 2 NO (g) + O 2(g) ∆H = +112kJ Note that the ∆H value must also be multiplied x2! Note that the ∆H value must also be multiplied x2! Net equation: Net equation: N 2 + O 2 2NO ∆H = +112kJ+ 68 kJ N 2 + O 2 2NO ∆H = +112kJ+ 68 kJ ANSWER =+ 180 kJ !