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Hess’ Law

Hess’ Law. Created by P. Perkerson. While we often represent chemical reactions with a single equation, it is important to understand that many reactions occur in a series of steps.

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Hess’ Law

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  1. Hess’ Law Created by P. Perkerson

  2. While we often represent chemical reactions with a single equation, it is important to understand that many reactions occur in a series of steps. • Enthalpy changes are not always rapid – some take millions of years. Think about diamond formation and rusting of iron. • Hess’ law says you can add heat from 2 or more thermo chemical equations to get the final equation and the final heat of reaction.

  3. Characteristics of ∆H • If a reaction is reversed the sign of ∆H is also reversed. • The magnitude of ∆H is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced equation are multiplied or divided by an integer then the ∆H will be multiplied or divided by the same integer. (page 336)

  4. Ex. C(s graphite) + O2(g)  CO2(g) ∆H = -394 kJ C(s diamond) + O2(g)  CO2 (g) ∆H = -396 kJ To determine change in enthalpy we will reverse the second equation which changes the sign of the ∆H. C(s graphite) + O2(g) CO2(g) ∆H = -394 kJ CO2(g) C(s diamond) + O2(g) ∆H = +396 kJ Cancel like terms and resulting equation is: C(s graphite)  C(s diamond) ∆H = 2 kJ

  5. C(s graphite)  C(s diamond) ∆H = 2 kJ • Is this reaction endothermic or exothermic? How do you know? • Let’s try another one! • Calculate ∆H for the reaction • S(s) + O2(g)  SO2(g) • S(s) + 3/2 O2(g)  SO3 (g) ∆H = -395.2 kJ • 2SO2(g) + O2(g)  2SO3(g) ∆H = -198.2 kJ

  6. Which equation do we need to reverse? • S(s) + 3/2 O2(g)  SO3 (g) ∆H = -395.2 kJ • 2SO2(g) + O2(g)  2SO3(g) ∆H = -198.2 kJ • Calculate ∆H for the reaction • S(s) + O2(g)  SO2(g) • We will also need to divide the second equation by 2. Why? • S(s) + 3/2 2/2 O2(g) SO3 (g) ∆H = -395.2 kJ • 2SO3(g)2SO2(g) + O2(g) ∆H = +198.2 kJ • SO3(g)SO2(g) + 1/2O2(g) ∆H = +99.1 kJ • Result: S(s) + O2(g)  SO2(g)

  7. Your turn! • Calculate ∆H for 2NO2(g)  N2O4(g) • N2(g) + 2O2 (g)  2NO2(g) ∆H = 67.7kJ • N2(g) + 2O2 (g)  N2O4(g) ∆H = 9.7 kJ

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