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Thermochemistry Section 3: Hess’ Law Hess’ Law since H is a state function, the change in H is independent of pathway Hess’ Law- when going from a set of reactants to a set of products, the ∆H is the same whether it happens in one step or a series of steps Example 1 Example 1 N2(g) + 2O2(g) 2NO2(g) ∆H = 68 kJ OR N2(g) + O2(g) 2NO(g) ∆H = 180 kJ 2NO(g) + O2(g) 2NO2(g) ∆H = -112 kJ N2(g) + 2O2(g) 2NO2(g) ∆H = 68 kJ Rules 1. If a reaction is reversed, the sign of ∆H must be reversed as well. – because the sign tells us the direction of heat flow as constant P 2. The magnitude of ∆H is directly proportional to quantities of reactants and products in reaction. If coefficients are multiplied by an integer, the ∆H must be multiplied in the same way. – because ∆H is an extensive property Example 2 Using the enthalpies of combustion for graphite (-394 kJ/mol) and diamond (-396 kJ/mol), find the ∆H for the conversion of graphite to diamond. Cgraphite (s) Cdiamond (s) ∆H=? Example 2 (1) Cgraphite(s) + O2(g) CO2(g) (2) Cdiamond(s) + O2(g) CO2(g) ∆H= -394kJ/mol ∆H= -396kJ/mol to get the desired equation, we must reverse 2nd equation: (1) Cgraphite(s) + O2(g) CO2(g) ∆H= -394kJ/mol (2) CO2(g) Cdiamond(s) + O2(g) ∆H= (-1)-396kJ/mol Cgraphite (s) Cdiamond (s) ∆H= -394 + 396 ∆H= 2 kJ/mol Example 3 Find ∆H for the synthesis of B2H6, diborane: 2B(s) + 3H2(g) B2H6(g) Given: (1) 2B(s) + 3/2O2(g) B2O3(s) ∆H1=-1273kJ (2) B2H6(g) + 3O2(g) B2O3(s) + 3H2O(g) ∆H2=-2035kJ (3) H2(g) + ½O2(g) H2O(l) (4) H2O(l) 3H2O(g) ∆H3=-286kJ ∆H4=44 kJ Example 3 Start by paying attention to what needs to be on reactants and products side (1) 2B(s) + 3/2O2(g) B2O3(s) ∆H1= -1273kJ (2) B2O3(s) + 3H2O(g) B2H6(g) + 3O2(g) ∆H2= (-1)-2035kJ (3) H2(g) + ½O2(g) H2O(l) ∆H3= -286kJ (4) H2O(l) 3H2O(g) ∆H4= 44 kJ Example 3 Underline what you want to keep- that will help you figure out how to cancel everything else: (1) 2B(s) + 3/2O2(g) B2O3(s) ∆H1= -1273kJ (2) B2O3(s) + 3H2O(g) B2H6(g) + 3O2(g) ∆H2=(-1)-2035kJ (3) H2(g) + ½O2(g) H2O(l) ∆H3= -286kJ (4) H2O(l) 3H2O(g) ∆H4= 44 kJ Example 3 Need 3 H2 (g) so 3 x (3) Need 3 H2O to cancel so 3 x (4) (1) 2B(s) + 3/2O2(g) B2O3(s) (2) B2O3(s) + 3H2O(g) B2H6(g) + 3O2(g) (3) 3H2(g) + 3/2O2(g) 3H2O(l) (4) 3H2O(l) 3H2O(g) ∆H1= -1273kJ ∆H2=(-1)(-2035kJ) ∆H3=(3)(-286kJ) ∆H4=(3)(44 kJ) 2B(s) + 3H2(g) B2H6(g) ∆H = -1273 + -(-2035) + 3(-286) + 3(44) = 36kJ