Enthalpy

EnthalpyEnthalpy

By: Veronica Mendez and By: Veronica Mendez and Quinn ShollenbergerQuinn Shollenberger

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DefinitionDefinition

EnthalpyEnthalpy- the heat content of a - the heat content of a system at constant pressure.system at constant pressure.

Enthalpy- ∆ HEnthalpy- ∆ H

Equation #1Equation #1

∆∆H = nCp∆TH = nCp∆T– nn is moles is moles

–CCp p is specific heat capacity is specific heat capacity (usually a given value)(usually a given value)

– ∆∆T T is the change in is the change in temperature temperature

Equation #2Equation #2

∆∆H =∑ ∆HH =∑ ∆Hp p - ∑ ∆H- ∑ ∆Hrr

– ∑ ∑ ∆∆HHpp is the summation of the products is the summation of the products

– ∑ ∆∑ ∆HHr r is the summation of the reactants is the summation of the reactants

Example problemExample problem

Find Find ΔΔH of the reaction.H of the reaction.– 1CH1CH44 + 2O + 2O22 --> 1CO --> 1CO22 + 2H + 2H22OO

– Mole Ratio is 1:2:1:2 because the Mole Ratio is 1:2:1:2 because the coefficient in front of CHcoefficient in front of CH4 4 is 1; is 1; coefficient in front of Ocoefficient in front of O2 2 is 2; coefficient is 2; coefficient in front of COin front of CO2 2 is 1; coefficient in front of is 1; coefficient in front of HH22O is 2.O is 2.

Example problem: Set up Example problem: Set up

ΔΔH = [H = [∑∑((ΔΔHHCO2CO2 + + ΔΔHHH2OH2O)] – )] – [[∑∑((ΔΔHHCH4CH4 + + ΔΔHHO2O2)])]

Example Problem: Plugging in Example Problem: Plugging in the numbersthe numbers

ΔΔH = [H = [(-393.5KJ/m X 1m) + (-286KJ/m (-393.5KJ/m X 1m) + (-286KJ/m X 2m)] – [(-74.8KJ/m X 1m) + 0][(-X 2m)] – [(-74.8KJ/m X 1m) + 0][(-393.5 KJ) + (-572 KJ)] –(-74.8 KJ)393.5 KJ) + (-572 KJ)] –(-74.8 KJ)

ΔΔH = -890.7 KJ (exothermic) H = -890.7 KJ (exothermic)

Example Problem: Further Example Problem: Further explanationsexplanations

The numbers from the equation:The numbers from the equation: ΔΔH = [H = [(-393.5KJ/m X 1m) + (-286KJ/m X 2m)] – [(-74.8KJ/m X 1m) (-393.5KJ/m X 1m) + (-286KJ/m X 2m)] – [(-74.8KJ/m X 1m)

+ 0][(-393.5 KJ) + (-572 KJ)] –(-74.8 KJ)+ 0][(-393.5 KJ) + (-572 KJ)] –(-74.8 KJ)

can be found on any Chemistry can be found on any Chemistry reference source. This includes CP reference source. This includes CP values.values.

Reminder

Do not forget to cross out your moles

Be careful with positive and negative signs

Quiz question 1

Calculate ΔΔH at 25H at 25°C for:°C for:

NaNa22O O (s)(s) + H + H22O O (g)(g) NaOH NaOH (s)(s)

Quiz question 2

If 15.0 g of Gold is heated from 16.1°C°C to 49.3°C, what is the heat °C, what is the heat absorbed by the gold?absorbed by the gold?

Quiz question 3

When 4.4 Kg of NaCl cools from 67.2°C°C to 25.0°C°C, how much heat is gained by the surroundings from the NaCl?

(Cp NaCl 50.5 J°C°C-1-1MM-1-1))

Quiz Question 4

Determine the standard molar heat of combustion (ΔΔH ) id Methanol, H ) id Methanol, CHCH33OH, when it is burned. OH, when it is burned. Use HUse H2200(g)(g)

Explanation to Quiz question 1

Steps:

•Balance equation: Na2O+H2O 2NaOH

•[(- 427KJ/m)(2m)]-[(242KJ/m)(1m) + (- 415.9KJ/m)(1m)]

• ΔΔH = -196.1KJH = -196.1KJ

Explanation to Quiz question 2

∆∆H = nCp∆TH = nCp∆T– 15.0 g(1m/ 197g)= 0.07614m15.0 g(1m/ 197g)= 0.07614m– (0.07614m)(25.4J/m°C)(33.2°C)(0.07614m)(25.4J/m°C)(33.2°C)– ∆∆H = 64.2 JH = 64.2 J

Explanation to Quiz Question 3

∆∆H = nCp∆TH = nCp∆T– Convert kg into g– 4400 g (1m/58.5g) = 75.2m– (75.2m)(50.5J/m°C)(42.2 °C) °C)(42.2 °C) – ∆∆H = 1.60 x 10 JH = 1.60 x 10 J

Explanation to Quiz question 4

Combustion reaction and balance the equation

2CH3OH + 3O2 2CO2 + 4H2O

Explanation for Quiz question 4 cont…

[(-393.5KJ/m)(2m)+(242KJ/m)(4m)] – [0+(-239KJ/m)(2m)]

∆∆H = 1277 KJH = 1277 KJ

The End!!!!