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Heat of reaction • Chemical reactions involve breaking of bonds (in reactants) and making of bonds (products). • At the end of the reaction, energy may be released to the surroundings or it may be absorbed from the surroundings. • A reaction that releases energy to its surroundings is called an exothermic reaction. • A reaction that absorbs energy from the surroundings is called an endothermic reaction. 15 Effects of heats of reactions Type of reaction Experimental effect What happens to the system Sign of “q” Endothermic Reaction vessel cools Energy is added + Exothermic Reaction vessel warms Energy is released - Ba(OH)2·8H2O(s) + 2NH4NO3(s) C3H8(g) + O2(g) 2NH3(g) + 10H2O(l) + Ba(NO3)2(aq) q = + 170.4 kJ CO2(g) + H2O(l) q = - 890 kJ 16 State functions • Any property that does not depend on the history of the system is called a state function. • The state of a system is defined as the chemical composition and its temperature, pressure and volume. • Pressure, temperature and volume are examples of state functions. • Changes in state functions are usually measured and are denoted by Δ followed by whatever state function is being measured. • The change in temperature ΔT can be measured by: ΔT = Tfinal - Tinitial 17 1 Enthalpy (H) & Enthalpy change (ΔH) • The heat absorbed or released by a system usually depends on the conditions under which the reaction is performed. • Normally, reactions are performed in vessels open to the atmosphere and hence at constant atmospheric pressure. • Enthalpy is an extensive property of a substance that can be used to obtain the heat evolved or absorbed in a chemical reaction. • Enthalpy is a state function. 18 Enthalpy of a reaction • State functions are properties that are independent of the previous history of the system. They depend on the present state of the system. • The ΔH for a reaction at a given temperature and pressure is obtained by: ΔH = H(products) – H(reactants) • The change in enthalpy in a chemical reaction is = heat of reaction at constant pressure: ΔH=qp 19 Thermochemical Equations • It is often useful to write the ΔH with the balanced equation. • A thermochemical equation is a balanced chemical equation (including physical states) with the ΔH written directly after the equation. 2H2(g) + O2(g) 2H2O(g) ΔH = -483.7 kJ 2H2(g) + O2(g) 2H2O(l) ΔH = -571.7 kJ NaHCO3(aq) + HCl(aq) NaCl(aq) + H2O(l) + CO2(g) ΔH = +11.8 kJ 20 2 Facts to Remember • For changes that are reverse of each other the ΔH values are numerically the same, but their signs are opposite. For the evaporation of water ΔH = +44.0 kJ/mol, while for condensation of water vapor, ΔH = 44.0 kJ/mol. • When a chemical equation is reversed the value of ΔH stays the same but the sign is reversed. N2(g) + 3H2(g) 2NH3(g) ΔH = -91.8 kJ 2NH3(g) N2(g) + 3H2(g) ΔH = +91.8 kJ • When a chemical equation is multiplied by any number the value of ΔH is also multiplied by that number. N2(g) + 3H2(g) 1/2N2(g) + 3/2H2(g) 2NH3(g) ΔH = -91.8 kJ NH3(g) ΔH = -45.9kJ Only when solving thermodynamics problems, can we have fractional coefficients 21 Stoichiometry and ΔH • How much heat is produced by the combustion of 1.00 lb (454 g) of propane (C3H8)? The ΔH for the combustion of propane is -2220 kJ mol-1. • Convert the grams of propane to moles of propane and then find the amount of heat produced: 22 Hess’s Law • The numerical value and the sign of the ΔH allows us to make informed decisions. • For several reactions a direct measurement can be done with a calorimeter. • Many times this is impossible or it is a time consuming task which makes it very hard. • Hess’s law allows us to manipulate equations for calculating ΔH for a given reaction. • If a reaction is the sum of two or more other reactions then the ΔH fro the overall process must be the sum of the constituent reactions. 23 3 Applying Hess’s law • What is the ΔH for the formation of methane from solid C and hydrogen gas? CH4(g) ΔH =? C(s) + 2H2(g) • Use the following information: ΔH (kJ) Reaction 1 2 3 C(s) + O2(g) CO2(g) H2(g) + 1/2O2(g) CH4(g) + 2O2(g) H2O(l) CO2(g) + H2O(l) -393.5 -285.8 -890.3 24 • The 3 equations as they are written cannot be added for the formation of CH4 from its elements. • We need the CH4 to be a product, but it is a reactant in equation 3. (Reverse equation 3, remember to change the sign also) • Equation (3’) needs 2 moles of H2O as reactants. Equation (2) is written for only one mole of H2O. Multiply equation (2) by 2 and multiply the ΔH value also by 2 to get: • After these modifications we can add the 3 equations to get the formation of CH4 from its elements. 25 ΔH (kJ) Reaction 1 C(s) + O2(g) 2’ 2H2(g) + O2(g) 3’ CO2(g) + 2H2O(l) C(s) + 2H2(g) CO2(g) 2H2O(l) CH4(g) + 2O2(g) CH4(g) -393.5 -571.6 + 890.3 - 74.8 26 4 Standard Enthalpies of Formation • The enthalpy change in a reaction in which reactants in their standard state produce products in their standard state is called the standard enthalpy of reaction (ΔH°). • The standard state of a substance is its pure and most stable form at 1 atmosphere pressure and at 25°C. • The enthalpy change for the formation of one mole of a compound in its standard state from its elements in their reference form and in their standard states is known as the enthalpy of formation ΔH°f of that compound. • The ΔH°f values (in appendix C in our book) can be used for calculating ΔH°. • The ΔH°f values for all elements are assigned a value = zero. 27 Using ΔH°f • Consider the reaction between methane and chlorine: CCl4(l) + 4HCl(g) ΔH° = ? CH4(g) + 4Cl2(g) • From Appendix C we can look up the values for the ΔH°f of CH4(g), CCl4(l) and HCl(g) and then write: ΔH°f (kJ) Reaction 1 C(graphite) + 2H2(g) CH4(g) -74.9 2 C(graphite) + 2Cl2(g) CCl4(l) -139 3 1/2H2(g) + 1/2Cl2(g) HCl(g) -92.3 28 • Apply Hess’s law. • Since the ____needs to be on the ____ and the ____ and _____ on the right, _______ equation _ and multiply equation _ by _ to get: ΔH°f (kJ) Reaction 1’ CH4(g) C(graphite) + 2H2(g) 2 C(graphite) + 2Cl2(g) 3’ 2H2(g) + 2Cl2(g) CH4(g) + 4Cl2(g) CCl4(l) 4HCl(g) CCl4(l) + 4HCl(g) The ΔH° for the reaction is ______ 29 5 Simplification • The previous problem can be greatly simplified by using the following formula: ΔH o = ∑ n ΔH (products) −∑ m ΔH (reactants) o f o f ΔH° = [ΔH°f (CCl4) + 4 ΔH°f (HCl)] – [ΔH°f (CH4) + 4 ΔH°f (Cl2) ΔH° = [(-139 kJ) + 4(-92.3 kJ)] – [(-74.9 kJ) + 4(0 kJ)] ΔH° = -433 kJ 30 Another example • Calculate the ΔH°vap of CS2 at 25°C. This is a phase change process. The vaporization process is given by: CS2(l) ΔH = o ∑n CS2(g) ΔH of (products) − ∑ m ΔH (reactants) o f • ΔH° = • ΔH° = 31 Another one • NH3 is used to prepare HNO3. Oxidation of NH3 to produce NO is the first step. What is the standard enthalpy change for this reaction? 4NH3(g) + 5O2(g) ΔH o = 4NO(g) + 6H2O(g) ∑ n ΔH (products) −∑ m ΔH (reactants) o f o f • ΔH° = • ΔH° = • ΔH° = 32 6 Calorimetry • A technique for measuring the heat evolved by a reaction. • A weighed sample is placed in a dish that is encased in a “bomb”. • The bomb is then placed in a well-insulated container. • The bomb is filled with pure O2 and then the sample is ignited. • The heat released by this burning raises the temperature of the water around it. • System = Sample + O2 • Surroundings = Bomb + water • Heat lost by system = heat qreaction = -(qwater + qbomb) gained by surroundings. 33 ΔH by Calorimetry • Octane is the primary component of gasoline. Suppose a 1.00 g sample of Octane (C8H18) is burned in a bomb calorimeter containing 1.20 kg of water. The temperature of the water and the bomb rise from 25.00°C to 33.50°C. If the heat capacity of the bomb = 837 J/K, calculate the amount heat transferred by the combustion of 1.00 g of octane. Also calculate the amount of heat transferred per mole. C8H18(l) + 25/2 O2(g) 8CO2(g) + 9H2O(l) qwater = (sp.heat of water) x (mass) x (ΔT) 34 • The heat released by the reaction also heats up the bomb and this is calculated by: • The total heat released by the reaction is calculated by: • This shows that the amount of heat released by the combustion of 1.00 g of Octane. • The amount of heat released by the combustion of 1 mole of Octane can be calculated by: 35 7 Coffee-cup Calorimetry • A bomb calorimeter is sometimes inconvenient to use, instead a coffee-cup calorimeter is used. • A coffee-cup calorimeter is simple, inexpensive and operates at constant (atmospheric) pressure. • The heat of reaction measured in a coffee-cup calorimeter is the measure of the reaction enthalpy (qp = ΔH). 36 Coffee-cup Calorimetry • 0.500g of Magnesium chips are placed in a coffee cup calorimeter containing 100.0 ml of 0.1M HCl. The temperature of the solution increases from 22.2°C to 44.8°C. What is the ΔH for this reaction per mole of Mg? Assume that the heat capacity of the solution is 4.20 J/gK and the density of HCl is 1.00 g/ml Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) 37 8