Stoichiometry: Real World Reactions
Heat of Reaction
Enthalpy is defined as how much heat a substance has at a given temperature and pressure, and is symbolized by the symbol H . This temperature and pressure is usually STP. Although there is no way to measure the absolute enthalpy of a substance, changes in enthalpy can be measured.
Heat of Reaction
The change in enthalpy for a reaction is known as the heat of reaction and has symbol δH . δH is negative for all exothermic reactions and positive for all endothermic reactions.
The states (gas, liquid, solid) of all products and reactants must be stated in any reaction involving enthalpy. Note how the state of the product in the following two reactions affects δH .
|H2 +1/2O2→H2O (gas) δH = - 241.8 kJ|
|H2 +1/2O2→H2O (liquid) δH = - 285.8 kJ|
Heat of Formation
The heat of formation is defines as the δH The next term you need to be familiar with is the standard heat of formation. It is defined as the δH for a reaction that produces 1 mole of compound from its constituent elements. It has its own special symbol, δH f . When solving enthalpy problems, you can find the heat of reaction using the following formula:
|δH = δH f (products) - δH f (reactants)|
Problem: Find δH for the reaction of sulfur dioxide with oxygen to form sulfur trioxide given the following heats of formation:
|δH fSO2 = - 296.8 kJ / mole|
|δH fSO3 = - 395.7 kJ / mole|
Solution: First write a balanced equation:
|2SO2(g) + O2(g)→2SO3(g)|
In this case, no conversion of moles or mole ratio is necessary. Based on the coefficients, simply assume 2 moles 2SO2(g) and 1 mole O2(g) . You want to find δH for the basic equation not a special case. You probably noticed that the δH f for O2 was not given. It is important to realize that the heat of formation for any element in its basic state is arbitrarily set to 0 kJ / mole. The diatomic molecules H2(g) , N2(g) , O2(g) , F2(g) , Cl2(g) , Br2(l) , and I2(s) are included in the grouping of elements in their fundamental states. Knowing this, let's find the components needed for the δH equation.
|δH (reactants) = + = - 593.6 kJ|
|δH f products = = - 791.4 kJ|
|δH = δH F (products) - δH F (reactants) = - 791.4 kJ - (- 593.6 kJ) = - 197.8 kJ|