Heat & Enthalpy Changes

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Presentation transcript:

Heat & Enthalpy Changes Objectives: 1. Define the terms enthalpy and enthalpy changes. 2. Explain what information is given by the standard enthalpy change of a reaction. 3. Explain how a calorimeter is used to determine the quantity of heat transferred in a chemical reaction. Key Terms: Enthalpy, standard enthalpy change, calorimetry

Calorimetry Calorimetry is the measurement of the heat gained or lost in a chemical reaction. A calorimeter is a device that isolates the outside environment. Any loss or gain of energy in the products results in a change in temperature only in the calorimeter. The rise (or drop) of temperature in the calorimeter is dependent on the amount of heat released or gained and the heat capacity of the surroundings Bomb calorimeter q = (masswater)(Cwater)(DT)

Enthalpy The difference in the energies related to products and reactants along is called enthalpy. Enthalpy is represented by the capital letter H DH = -qenvironment DH = Hproducts – Hreactants if exothermic then enthalpy is negative (- DH) if endothermic then enthalpy is positive (+DH) Standard Enthalpy change (DHo) occurs at constant temperature (25oC) and pressure (1atm) Represents a single mole Is calculated using the most stable allotrope of an element

Interpreting enthalpy equations C(s) + H2O(l) + 113kJ g CO(g) + H2(g) DHo = + 113kJ In this reaction 113kJ of heat energy is absorbed in the endothermic reaction. Since the products have more energy than the reactants the products are said to have undergone positive enthalpy (a positive gain in chemical potential energy). endothermic reactions result in positive enthalpy (+H)

Interpreting enthalpy equations C3H8(g) + 5O2(g) g 3CO2(g) + H2O(g) + 2043kJ DHo = - 2043kJ In this reaction 2043kJ of heat energy is released in the exothermic reaction. Since the products have less energy than the reactants the products are said to have undergone negative enthalpy (a negative gain in chemical potential energy). exothermic reactions result in negative enthalpy (-H)

Solving Enthalpy Consider this question. C3H8(g) + 5O2(g) g 3CO2(g) + H2O(g) + 2043kJ DHo = - 2043kJ If 2.50 moles of oxygen are reacted with 1.00 mol of C3H8, how much heat is produced in the reaction? What is the enthalpy? (2.5/5) x 2043kJ = qrxn DH = -q Since you start out with 1/2 the quantity of O2 required for the reaction, the heat produced is also halved. (1021.5kJ) DH = -1020kJ Ratio of coefficients

Using Calorimetry to solve enthalpy Examine the reaction of sodium hydroxide water NaOH(s) g Na+(aq) + OH-(aq) 0.050 moles of sodium hydroxide reacts in a calorimeter containing 75.0g of water. The temperature of the water is raised from 19.8oC to 26.7oC. What is the enthalpy of the rxn? q = (masswater)(Cwater)(DT) qNaOH = (75g )x(4.184J)/goC)x(26.7oC-19.8oC) = +2170J Denthalpy (DH)= -Denergy (-Dq) Enthalpy (remember that the sign must change for enthalpy) DH = -2.170kJ Since the water in the calorimeter represents the environment, the reaction is exothermic DHo = (1 mol NaOH / 0.050 mol NaOH) x -2.17kJ = -4.3kJ