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ENTHALPY, HESS’ LAW, AND THERMOCHEMICAL EQUATIONS

ENTHALPY, HESS’ LAW, AND THERMOCHEMICAL EQUATIONS

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ENTHALPY, HESS’ LAW, AND THERMOCHEMICAL EQUATIONS

Quick Review of Concepts

We have been introduced to heat producing (exothermic) reactions and heat using (endothermic) reactions

Heat is a measure of the transfer of energy from a system to the surroundings and from the surroundings to a system

Enthalpy

The heat given off or absorbed during a reaction (change in heat of a system) is called the change in enthalpy (ΔH) when the pressure of the system in kept constant

DH = H (products) – H (reactants)

Hproducts < Hreactants

DH < 0

Hproducts > Hreactants

DH > 0 6.3

More on Enthalpy

Calorimetry

We measure the transfer of heat (at a constant pressure) of a chemical reaction by a technique called calorimetry

In calorimetry: The heat released by the system is equal to

the heat absorbed by its surroundings The heat absorbed by the system is equal

to the heat released by its surroundings The total heat of the system and the

surroundings remains constant First Law of Thermodynamics

Calorimetry

We use an insulated device called a calorimeter to measure this heat transfer

A typical device is a “coffee cup calorimeter” Reaction is open to the atmosphere

Therefore, constant pressure q = ∆H at constant pressure

Calorimetry

To measure ΔH for a reaction using a coffee-cup calorimeter:1.Dissolve the reacting

chemicals in known volumes of water

2.Measure the initial temperatures of the solution

3.Mix the solutions4.Measure the final temperature

of the mixed solution

Calorimetry

The heat generated by the reactants is absorbed by the water

We know the mass of the water, mwater

We know the change in temperature, ∆Twater

We also know that water has a specific heat of Cwater = 4.18 J/°C-g.

We can calculate the heat of reaction by:qsys = ∆H = −qsurr

qrxn = ∆H = −qwater

qrxn = ∆H = -(mwater × Cwater × ∆Twater)

Practice Problem

• When 25.0 mL of water containing 0.025 mol of HCl at 25.0°C is added to 25.0 mL of water containing 0.025 mol of NaOH at 25.0°C in a coffee cup calorimeter, a reaction occurs. Calculate ∆H (in kJ) of one mole of water during this reaction if the highest temperature observed is 32.0°C.

• Note:

• Assume the densities of the solution are 1.00 g/mL

• Assume all heat was absorbed by water

Summary

Heat is a measure of the transfer of energy from a system to the surroundings and from the surroundings to a system

The change in heat of a system is called the change in enthalpy (ΔH) when the pressure of the system in kept constant

We measure the transfer of heat (at a constant pressure) of a chemical reaction by a technique called calorimetry

We use an insulated device called a calorimeter to measure this heat transfer

H2O (s) H2O (l) DH = + 6.01 kJ

Is DH negative or positive?

System absorbs heat

Endothermic

DH > 0

6.3

Thermochemical Equations

Is DH negative or positive?

System gives off heat

Exothermic

DH < 0

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890.4 kJ

Thermochemical Stoichiometry Enthalpy is STOICHIOMETRIC!

Recall that the stoichiometric coefficients always refer to the number of moles of a substance

You would read the above equation as:“6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm”

“890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm”

H2O (s) H2O (l) DH = 6.01 kJ

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890.4 kJ

6.3

How much heat is evolved when 266 g of white phosphorus (P4) burn in air?

P4 (s) + 5O2 (g) P4O10 (s) DH = -3013 kJ

266 g P4

1 mol P4

123.9 g P4

x 3013 kJ1 mol P4

x = 6470 kJ

Practice Problem

Let’s Try Some More!

Page 165

• If you reverse a reaction, the sign of DH changes

H2O (l) H2O (s) DH = -6.01 kJ

• If you multiply both sides of the equation by a factor n, then DH must change by the same factor n

2H2O (s) 2H2O (l) DH = 2 x 6.01 = 12.0 kJ

Manipulating ∆H