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Presentation transcript:

Bell Work. Take out a piece of paper Bell Work!!! Take out a piece of paper. You have 5ish mins to do the following problems: A certain mass of water was heated with 41, 840J, raising its temperature from 22.OoC to 28.5oC. Find the mass of the water, in grams. (specific heat of water =4.184 J/g0C) How much energy must be absorbed by 20.o grams of water to increase its temperature from 283.ooC to 303.o0C? (specific heat of water= 4.184 J/g0C) Convert 230C into Kelvin.

Enthalpy

Enthalpy vs. Entropy Enthalpy Entropy Specific Heat Hess’ Law 1st law of Thermodynamics Dispersal of energy

Specific Heat Definition: The amount of energy required to change the temperature of one gram of a substance by one Celsius degree Q=smΔT

Practice Problem A 1.6g sample of a gold requires 5.8J of energy to change its temperature from 23oC to 41oC. What is the specific heat of gold? 5.8 J = 1.6 g (x) (41-23) 5.8 J = 1.6 g (x) (18oC) 5.8 J = 28.8g0C (x) 0.201 J/g0C

Hess’s Law In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. So in other words, your going to find ΔH

Example Enthalpy change will equal ΔH1: N2 (g) + 2O2 (g)  2NO2 (g) ΔH1= 68 2 steps---- N2 (g) + O2 (g)  2NO (g) ΔH2= 180 2NO (g) + O2 (g)  2NO2 (g) ΔH3= -112 N2 (g) + 2O2 (g)  2NO2 (g) ΔH2 +ΔH3 = 68

If the reaction is reversed or needs to be reversed the sign of ΔH is also reversed If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer

Practice Problem Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: H2SO4 (l)  SO3 (g) + H2O (g) H2S (g) + 2O2 (g)  H2SO4 (l) ΔH=-235.5 kJ H2S (g) + 2O2 (g)  SO3 (g) + H2O (l) ΔH=-207 kJ H2O (l)  H2O (g) ΔH=44kJ

Practice Problem Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: 2C2H4O(l) + 2H2O(l) → 2C2H6O(l) + O2(g) C2H6O(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ΔH = -685.5 kj C2H4O(l) + 5/2O2(g) → 2CO2(g) + 2H2O(l) ΔH =-583.5 kJ

1st law of thermodynamics The total amount of mass and energy is constant. It is neither created nor destroyed. Law of conservation of mass and energy.