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CHEMISTRY 161 Chapter 6 www.chem.hawaii.edu/Bil301/welcome.html.

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Presentation on theme: "CHEMISTRY 161 Chapter 6 www.chem.hawaii.edu/Bil301/welcome.html."— Presentation transcript:

1 CHEMISTRY 161 Chapter 6 www.chem.hawaii.edu/Bil301/welcome.html

2 THERMODYNAMICS HEATCHANGE quantitative study of heat and energy changes of a system the state (condition) of a system is defined by T, p, n, V, E CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O(l)

3 the state (condition) of a system is defined by T, p, n, V, E STATE FUNCTIONS properties which depend only on the initial and final state, but not on the way how this condition was achieved

4 ΔV = V final – V initial Δp = p final – p initial ΔT = T final – T initial ΔE = E final – E initial

5 Energy is a STATE FUNCTION IT DOES NOT MATTER WHICH PATH YOU TAKE ΔE = m g Δh

6 CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) ENTHALPY, H Reactants Products CO 2 (g) + 2H 2 O(g) - 802 kJ - 88 kJ - 890 kJ Hess Law

7 heat is spontaneous transfer of thermal energy two bodies at different temperatures T 1 > T 2 spontaneous T1T1 T2T2 Zeroth Law of Thermodynamics a system at thermodynamical equilibrium has a constant temperature

8 First Law of Thermodynamics energy can be converted from one form to another, but cannot be created or destroyed CONSERVATION OF ENERGY

9 SYSTEM SURROUNDINGS THE TOTAL ENERGY OF THE UNIVERSE IS CONSTANT ΔE = ΔE system + ΔE surrounding = 0 - +

10 ΔE system = ΔQ + ΔW First Law of Thermodynamics ΔQ heat change ΔW work done  Q > 0 ENDOTHERMIC  Q < 0 EXOTHERMIC ?

11 ΔW = - p ΔV mechanical work M the energy of gas goes up M the energy of gas goes down ΔV < 0 ΔV > 0

12 First Law and Enthalpy ΔE system = ΔQ - pΔV 1.constant pressure → enthalpy change Δ H 2. ideal gas law → p V = n R T p ΔV = Δn R T ΔE system = ΔQ – R T Δn ΔQ = ΔE system + R T Δn = ΔH

13 Δ n = n final – n initial Calculate the energy change of a system for the reaction process at 1 atm and 25C 2 CO(g) + O 2 (g) → 2 CO 2 (g) ΔH o = -566.0 kJ ΔE system = ΔH 0 - R T Δn ΔE system = -563.5 kJ

14 A gas is compressed in a cylinder from a volume of 20 L to 2.0 L by a constant pressure of 10 atm. Calculate the amount of work done on the system.

15 Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 30.0°C. Zn(s) + 2H + (aq) → Zn 2+ (aq) + H 2 (g)

16 The heat of solution of KCl is +17.2 kJ/mol and the lattice energy of KCl(s) is 701.2 kJ/mol. Calculate the total heat of hydration of 1 mol of gas phase K + ions and Cl – ions.

17 Homework Chapter 6, problems


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