lecture 14 - Thermochemistry Many chemical reactions...

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Unformatted text preview: Thermochemistry Many chemical reactions are accompanied by heat release. In others, the chemicals tend to get colder so one has to provide heat to maintain the temperature. How can one predict these things? For example, consider the reaction C6H12O6 + 6O2 → 6CO2 + 6 H2O, which happens in your body every time you eat chocolate. Our organism uses sugar as “fuel”. But you can also burn sugar (in fact, the “calorie content” you find on food packages is measured by burning the food) – this is obviously an exothermic reaction. We’ll study a simpler example: CH4 + 2O2 = CO2 + 2H2O In this reaction a lot of heat is released. This can be used to power an engine or to heat up your house. Since heat is not a function of state, in order to quantify the “heat” of this reaction, we need to specify the process we want to talk about. That is, to specify its initial state, final state, and the path. Chemists use the following convention: Initial state: 1 mole of CH4, 2 moles of O2 , at given P and T. The reactants are separated (mot mixed) Final state: a mole of CO2 and 2 moles of H2O at given P and T. The products are separated. The reaction is complete (no reactants left). Path: constant P and T. The heat q exchanged between the system and its surroundings in this process is the heat of reaction. We know that at constant P and T: q = ΔH and that H is a function of state (i.e., independent of the path). This is also referred to the “enthalpy” of reaction, although the proper term should be the “change of enthalpy” in the reaction. If q<0 the heat is released and the reaction is exothermic. Otherwise it is endothermic. Clearly, the heat of the reverse reaction is –ΔH if the heat of the direct reaction is ΔH. Note 1. Heat of mixing & unmixing is not included here. Note 2. In tables, you’ll find “standard” heats of reactions (at P = 1 atm and T = 298.15K, or P= 1 bar T= 292.15K etc., depending what convention is used for the standard state). Some of reactions DO NOT take place under these conditions. The how were the heats measured? They were not. They were measured at different conditions and then calculated for different P and T. Thermochemistry does not care if the reaction really takes place in Nature. Kinetics is concerned with this issue. Kinetics tells us that some reactions are way too slow to observe under certain conditions. In the following, we will find out how to calculate the reaction heat at given temperature and pressure if it is known at some other values of P and T. Heats of formation. We could tabulate the heat of every possible reaction. Unfortunately, there are two many of them and some of them are difficult and/or expensive to carry out. Fortunately, thermodynamics tells us that in order to find the heat of a chemical reaction, we do not necessarily have to measure it. One method is based on using the heats of formation. The heat of formation of a compound is the heat of the reaction, in which a mole of the compound is formed from the elements, in their stable form. For example, the heat of formation of methane at room temperature and T = 1 atm is the heat of the reaction C(gr) + 2H2(g) → CH4 where “gr” stands for graphite. It is not the heat of the reaction C(diamond) + 4H→ CH4 because under these conditions the stable form of C is not diamond and hydrogen is a diatomic gas, not a monoatomic gas. We denote the heat of formation of methane: ΔHf(CH4) Consider now two different paths that convert CH4(g) + 2O2(g) into CO2(g)+ 2H2O(l) : Path 1: CH4(g) + 2O2(g) → CO2(g)+ 2H2O(l) Path 2: CH4(g) + 2O2(g) → C(gr) + 2H2(g) + 2O2(g) → CO2(g)+ 2H2O(l) step 1 step 2 Path 1 is the straightforward burning of methane. In path 2 we first decompose the reactants into the elements (in their stable state) (step1) and then assemble them into the products (step 2) . The heat of reaction in step 1 is ΔH(step 1 ) = –ΔHf(CH4(g)) – 2 ΔHf(O2(g)) . Notice that ΔHf(O2(g)) = 0. The heat of reaction in step 2 is: ΔH(step 2 ) = ΔHf(CO2(g)) + 2 ΔHf(H2O(l)) The total heat for path 2 is ΔH(step 1 ) + ΔH(step 2 ) = ΔHf(CO2(g)) + 2 ΔHf(H2O(l)) –ΔHf(CH4(g)) – 2 ΔHf(O2(g)) It easy to see what the general rule is: ΔH = So if we know the heat of formation for each reactant and product, we can compute the heat of reaction. Then, using the thermodynamic relations that we derived, we can also calculate the heat of reaction at any P and T. For the reaction CH4(g) + 2O2(g) = CO2(g)+ 2H2O(l), let us calculate ΔH at T=298K and P = 1 atm. Using the values of the heats of formation ΔHf that you can find in in any handbook of thermodynamic data (or at www.nist.gov) we find: ΔH = ΔHf(CO2(g)) + 2 ΔHf(H2O(l)) –ΔHf(CH4(g)) – 2 ΔHf(O2(g)) =  ­890.5 kJ/mol (*) Suppose now that we would like to know ΔH fore the same reaction at T= 328 K and P = 1atm. To do so, we write ΔH =  ­ H (CH4(g))  ­ 2 H (O2(g)) + H (CO2(g))+ 2 H (H2O(l)) (**) Notice that although the equations (*) and (**) look alike, it IS NOT TRUE that ΔHf(CO2(g)) = H (CH4(g)) etc. The heat of formation of a substance is a change of enthalpy in a particular reaction. It is not the same as the molar enthalpy of the same compound. This is very important! products ∑ ΔH f − reac tan ts ∑ ΔH f !H f ( A) " H ( A)! For a change of H (A) from state 1 to state 2 we have: ⎛ ∂V ⎞ H (2)  ­ H (1) = ∫ C p dT + ∫ ( V  ­ T ⎜ ⎟ )dP ⎝ ∂T ⎠ P 1 1 This provides us with a method of calculating Δ H for any P and T if we know it at some reference temperature and pressure (e.g., standard conditions). 2 2 Let us apply this result to our reaction. Since P is the same, this gives 328 H (328 K ) = H (298 K ) + 298 ∫ dTC P If we further assume that the heat capacity is temperature independent (reasonable if the temperature change is not very large) then H (328K ) ; H (298K ) + CP (328K − 298K ) Applying this relation for each reactant and product, we find ΔH(328K) =  ­ H (CH4(g), 298K)  ­ 2 H (O2(g), 298K) + H (CO2(g), 298K)+ 2 H (H2O(l), 298K) + ( ­ CP (CH4(g))  ­ 2 CP (O2(g)) + CP (CO2(g))+ 2 CP (H2O(l))) (328K ­298K) = ΔH(298K) + ( ­ CP (CH4(g))  ­ 2 CP (O2(g)) + CP (CO2(g))+ 2 CP (H2O(l))) (328K ­298K) The heat capacity values can again be found in thermochemistry reference books: CP (CH4(g))=35.7 J/mol K CP (O2(g)) = 29.4 J/mol K CP (CO2(g)) = 37.1 J/mol K CP (H2O(l))) = 75.3 J/mol K This finally gives ΔH(328K) =  ­887.7 kJ/mol More examples of using thermochemistry Example 1. Calculate ΔH and ΔS for the process applied to one mole of water at P = 1 atm: H2O(s,  ­10C) → H2O(l, +10C) The melting point of ice at P = 1 atm is 00C and ΔHmelting = 6 kJ/mol. Use C p (H2O(s)) = 36.8 J/(K mol) and C p (H2O(l)) = 75.3 J/(K mol) and assume that these heat capacities are temperature independent. Solution (don’t really need to know any thermochemistry): ΔH = C p (H2O(s)) (273K – 263K) + ΔHmelting + C p (H2O(l)) (283K ­273K) = 7121 J ΔS = C p (H2O(s)) ln(273K/263K) + ΔHmelting /273K + C p (H2O(l)) ln(283K/273K) = 26 J/K ...
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This note was uploaded on 04/24/2011 for the course CH 52635 taught by Professor Makarov during the Spring '11 term at University of Texas.

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