molar heat presentation

7/25/2019 Molar Heat Presentation

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Measuring Quantity of Heat

Module: Water

Measuring Quantity of Heat 1 / 24
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Heat Absorbed By The Environment

1

Calculate the amount of heat required to heat 155g from 15.0o

C to33.0oC.

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Qwater =mwaterCT

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Qwater =mwaterCT

The sign of the quantity of heat is positive since the sign is dependant onT

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Qwater =mwaterCT

The sign of the quantity of heat is positive since the sign is dependant onT

Tf>Ti

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Qwater =mwaterT (1)



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Qwater =mwaterCT

= (155g)(4.18J/g/oC)(33.0oC 15.0oC)

= 1.17 104J

(2)

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Molar Heat of Solution

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The Molar Heat of Solution (Hsoln) of a substance is the amount of

heat energy absorbed when one mole of a substance dissolves inexcess water.

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The Molar Heat of Solution (Hsoln) of a substance is the amount ofheat energy absorbed when one mole of a substance dissolves in

excess water.Remember:

Hsoln =HProducts HReactants

Where by convention HProducts


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The Molar Heat of Solution (Hsoln) of a substance is the amount ofheat energy absorbed when one mole of a substance dissolves inexcess water.

Remember:Hsoln =HProducts HReactants



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The Molar Heat of Solution (Hsoln) of a substance is the amount ofheat energy absorbed when one mole of a substance dissolves inexcess water.

Remember:Hsoln =HProducts HReactants



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Process Of Calculating The Molar Heat of Solution

We assume the dissolution reaction to be exothermic. As an equation:

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Qsolution = mCT



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Qsoln =

msolnCT

The molarheat of solution is then calculated as the energy released permoleof the substance

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Qsoln = msolnCT

The molarheat of solution is then calculated as the energy released permoleof the substance:

Hsoln = Qsoln

nsubstance


E l Q i


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Example Question

A calorimeter contained 50.0g of water. When 1.0g of dry ammonium

chloride powder was dissolved in water, the mixture dropped intemperature from 21.3oC to 21.3oC.


E l Q i
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Example Question

A calorimeter contained 50.0g of water. When 1.0g of dry ammoniumchloride powder was dissolved in water, the mixture dropped intemperature from 21.3oC to 20.1oC.

Assume that no energy was lost to the environment, calculate the molarenthalpy of solution for this dissolution reaction (Assume C = 4.2J/g/oC)


E l Q i
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Example Question

Data:

mwater = 50.0g

mNH4Cl= 1.0g

T = 20.1oC 21.3oC = 1.2oC4.2J/g/oC


E l Q ti
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Example Question

Qsoln = msolnCT

=

(mwater+ mNH4Cl)CT= (51.0g)(4.2J/g/oC)(1.2oC)

= 257.04J

(3)


E l Q ti
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Example Question

Hsoln = Qsoln

n(NH4Cl)

= 257.04J1.0g

53.492mol

= +13753J/mol

(4)


Proper Understanding Of The Formula
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Consider the fact that no heat is lost during the dissolution process. i.e.

Qlost+ Qgained= 0

Qreaction+ Qcalorimeter = 0

We note that reaction refers to the dissolution of the salt (which is still areaction)

Qreaction= Qcalorimeter


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Qreaction= Qcalorimeter

Now, Qcalorimeterrefers to the heat absorbed by the calorimeter.

Qreaction = mcalCT

But, mcal refers to the mass of the solution.

Qreaction= msolutionCT

And Qsolution =Qreaction

Qsolution = msolutionCT


molar heat presentation

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