Enthalpy Chapter 5.3-4,7

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Enthalpy Chapter 5.3-4,7 heat transfer in (endothermic), +q heat transfer out (exothermic), -q SYSTEM E = q + w w transfer in (+w) w transfer out (-w) Internal Energy at Constant Volume E = E K + E P ΔE = q + w Because most systems, especially those not involving gases, undergo little to no change in volume at the constant pressure of the atmosphere, the work that is done on or by the chemical system is small. When the change in volume of a system is zero, work energy lost or gained is also zero: w = -P external ΔV So at constant volume, the change in internal energy is equal to heat gained or lost: Enthalpy Since the volume change accompanying many reactions is close to zero, another thermodynamic quantity is often used, called enthalpy. Enthalpy is the thermodynamic quantity equivalent to the total heat content of a system. Enthalpy is defined as the internal energy plus the product of pressure and volume: H = E + PV ΔE = q v Enthalpy H = E + PV When the system changes at constant pressure, the change in enthalpy, H, is This can be written H = (E + PV) H = E + P V In this equation, the change in total heat content of the system is equal to the sum of E, thought as the total the energy required to make the system, and P V, thought as the total energy needed to make room for the system. Enthalpy When at constant pressure, the H value assumes no work energy is lost or gained other than expansion work that makes room for the system or from the atmosphere collapsing. Since E = q + w and w = P V, we can substitute into the change in enthalpy expression: H = E + P V H = (q + w ) + P V H = (q + ( P V)) + P V H = q p

Enthalpy vs. Internal Energy At constant pressure the observed heat gained or lost by the system is the change in enthalpy H = q p At constant volume the observed heat gained or lost by the system is the change in internal energy : ΔE = q v Given the volume change of many reactions is close to zero, the difference between H and ΔE is small; making H a satisfactory measure of energy changes during most chemical reactions. Types of Enthalpy Changes Enthalpy changes ( H) can be observed in one of three modes: 1) Heat is transferred from system into surroundings, or visa versa, with no physical or chemical change. ) Heat is transferred as the system or the surroundings undergoes a physical change (phase change, dissociation, etc ) 3) Heat is transferred between the system (reactants and products) and the surroundings as a chemical change occurs. Important Slide! Enthalpy Change of a Chemical Reaction All chemical reactions either release or absorb heat, even if we can not observe it directly. Reactants Products + Energy (exo) Learning Check Identify each reaction as (Ex) exothermic or (En) endothermic. A. N (g)+ 3H (g) NH 3 (g) + 9 kj B. CaC 3 + 556 kj Ca + C (g) C. S (g) + (g) S 3 (g) + heat Reactants + Energy Products (endo) 10 Basic Chemistry Copyright 011 Pearson Education, Inc. The change in enthalpy can be described by: ΔH = H f H i The total enthalpy of the products minus the total enthalpy of the reactants describes the total change in internal heat energy of a chemical reaction. The enthalpy change for any given reaction is called the reaction enthalpy (ΔH rxn ) Example: The combustion of hydrogen gas with oxygen to produce water produces 483.6 kj of heat energy Since energy was produced, the reaction is said to be exothermic. H + H (g) + 483.6kJ

Endothermicity and Exothermicity A process is endothermic when H is positive. A process is exothermic when H is negative. Two ways to express energy transfer in thermochemical equations: S + S 3(g) +197.8 kj S + S 3(g) H = -197.8 kj Notice, when the enthalpy is represented after the equation the sign changed; the reaction is exothermic Write the thermochemical equations for the combustion of methane The heat released or absorbed by a reaction depends on the physical states of the reactants (s, l or g) and on the physical environment (T and P). bserve the difference in the following: CH 4(g) + (g) C + H (g) ΔH = -80 kj CH 4(g) + (g) C + H (l) ΔH = -890 kj The Truth about Enthalpy 1. Enthalpy is an extensive property.. H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction. 3. H for a reaction depends on the state of the products and the state of the reactants. Because thermochemical reactions are affected by physical states and environmental conditions, standard conditions must be defined to compare enthalpies of different reactions. Standard Molar Enthalpy of a Reaction ΔH o rxn = the enthalpy of a reaction in which all reactants and products are in their standard states and 1 atm pressure. The standard state refers to the physical state of a pure substance at 1 atm.

So, which reaction indicates the enthalpy of the reaction under standard state conditions? CH 4(g) + (g) C + H (g) ΔH = -80 kj CH 4(g) + (g) C + H (l) ΔH = -890 kj It is understood that the coefficients of a balanced chemical reaction describes the molar ratio of reactants and products. The coefficients for a chemical reaction also describe the ratio of heat produced/consumed by any chemical reaction. Notice in the standard state combustion of methane: CH 4(g) + (g) C + H (l) ΔH = -890 kj For every one mole of methane combusted, 890 kj of energy are produced. What about: 1 mol water? 4 mol oxygen? Calculations with CH 4(g) + (g) C + H (l) ΔH = -890 kj 1. Calculate the amount of energy that is released when.99 grams of methane gas is reacted with excess oxygen. Calculations with N + + 181kJ N (g). If 15.0 g of N is produced, how many kj was absorbed? Calculations with S + S 3(g) +197.8 kj S + S 3(g) H = -197.8 kj 3. Calculate the energy produced from the reaction as.00 L of sulfur dioxide gas reacts with.5 mol of oxygen gas at STP.

For any reaction, an energy diagram (enthalpy diagram) can be drawn: For the reaction: H + H (g) + 483.6kJ The enthalpy diagram for the reaction is: H H + H (g) ΔH < 0 (exo) = 483.6 kj H + ΔH = 483.6 kj H (g) 4. Write the correct thermochemical equation for a reaction that requires 51KJ of energy to combine xenon gas with fluorine gas producing xenon tetraflouride. Then generate an enthalpy diagram for the reaction. I find your lack of understanding disturbing Reaction Enthalpies from Standard Enthalpies of Formation Standard Enthalpy of Formation The change in enthalpy that accompanies the formation of one mole of a compound from its free elements in their standard states. 1/N + N ΔH o f =34 kj/mol Notice: ΔH o f is only for the production of one mole of a substance ΔH o f is zero for any pure element in its standard state Standard Enthalpies of Formation Standard enthalpies of formation, H f, are measured under standard conditions (5 C and 1.00 atm pressure). Examine the following reaction for the combustion of methane CH4(g) (g) C H (l) H rxn =?

For the reaction to occur, reactants must be broken apart and the products must form. 1. C H CH 4(g) ΔH o f =-75 kj/mol. C C Δ H o f =-394 kj/mol 3. H 1 H (l) Δ H o f =-86 kj/mol CH C H 4(g) C H C H (l) - ΔH ΔH ΔH 1 3 Theoretical Calculation of H For any chemical reaction, Enthalpies of Formation can be used in this way to find the Enthalpy change under standard state conditions: ma + mb nc + nd CH ΔH 4(g) rxn ( ΔH ) ΔH 1 C H H 3 (l) H = n H f,products m H f,reactants where n and m are the stoichiometric coefficients. 5. Calculate the enthalpy change for the following reactions: A. Combustion of solid glucose (C 6 H 1 6 ) B. Methane reacting with chlorine to form methyl chloride, CH 3 Cl. Now, combining what we know: 6. Calculate the amount of energy consumed or released as the following quantities react. A..34 g of ethane, (C H 6 ), combusts in excess oxygen. B. 10.0 g of sodium metal reacts to with 10.0 g of liquid hydrogen peroxide to form solid sodium hydroxide.