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

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

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)

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

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

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

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

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

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

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

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

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

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

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

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.

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)

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.

Homework Chapter 6, problems