Hess Law Yes

7/31/2019 Hess Law Yes

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Lecture : Hess Law

Outline

Definition of Hess Law

Using Hess Law (examples)


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First Law of Thermodynamics

First Law: Energy of theUniverse is Constant

E = q + w

q = heat. Transferredbetween two bodies

w = work. Force acting overa distance (F x d)

wq

EEE initialfinalsys


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Definition of Enthalpy

Thermodynamic Definition of Enthalpy (H):

H = E + PV

E = energy of the system

P = pressure of the system

V = volume of the system


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Changes in Enthalpy Consider the following expression for a chemical

process:

DH = Hproducts - Hreactants

IfDH >0, then qp >0. The reaction is endothermic

IfDH


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Thermodynamic State Functions

Thermodynamic State Functions:

Thermodynamic properties that are dependent on

the state of the system only. (Example: DE andDH)

Other variables will be dependent on pathway

(Example: q and w). These are NOT state

functions. The pathway from one state to the other

must be defined.


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Hess Law Defined

Enthalpy is a state function. As such, DH for

going from some initial state to some final state is

pathway independent.

Hess Law: DH for a process involving the

transformation of reactants into products is not

dependent on pathway. Therefore, we can pick

any pathway to calculate DH for a reaction.


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Hess Law: An Example


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Using Hess Law

When calculating DH for

a chemical reaction as a

single step, we can use

combinations of

reactions as pathways

to determine DH for our

single step reaction.

2NO2 (g)

N2 (g) + 2O2 (g)

q

2NO2 (g)N2 (g) + 2O2 (g)


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Example (cont.)

Our reaction of interest is:

N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ

This reaction can also be carried out in two

steps:

N2 (g) + O2 (g) 2NO(g) DH = 180 kJ

2NO (g) + O2 (g) 2NO2(g) DH = -112 kJ


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Example (cont.)

If we take the previous two reactions and add

them, we get the original reaction of interest:

N2 (g) + O2 (g) 2NO(g) DH = 180 kJ

2NO (g) + O2 (g) 2NO2(g) DH = -112 kJ

N2 (g) + 2O2 (g) 2NO2(g) DH = 68 kJ


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process:



IfDH


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Example (cont.)

Note the important things about this example, the

sum ofDH for the two reaction steps is equal to

theD

H for the reaction of interest.

We can combine reactions of known DH to

determine the DH for the combined reaction.


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Hess Law: Details

Once can always reverse the direction of a

reaction when making a combined reaction.

When you do this, the sign ofDH changes.

N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ

2NO2(g) N2(g) + 2O2(g) DH = -68 kJ


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Details (cont.)

The magnitude ofDH is directly proportional tothe quantities involved (it is an extensive

quantity).

As such, if the coefficients of a reaction are

multiplied by a constant, the value ofDH is also

multiplied by the same integer.

N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ

2N2(g) + 4O2(g) 4NO2(g) DH = 136 kJ


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Using Hess Law

When trying to combine reactions to form a

reaction of interest, one usually works

backwards from the reaction of interest. Example:

What is DH for the following reaction?

3C (gr) + 4H2 (g) C3H8 (g)


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Example (cont.)

3C (gr) + 4H2 (g) C3H8 (g) DH = ?

Youre given the following reactions:

C (gr) + O2 (g) CO2 (g) DH = -394 kJ

C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) DH = -2220 kJ

H2 (g) + 1/2O2 (g) H2O (l) DH = -286 kJ


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Example (cont.)

Step 1. Only reaction 1 has C (gr).

Therefore, we will multiply by 3 to get the

correct amount of C (gr) with respect to ourfinal equation.

C (gr) + O2 (g) CO2 (g) DH = -394 kJ

3C (gr) + 3O2 (g) 3CO2 (g) DH = -1182 kJ

Initial:

Final:


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Example (cont.)

Step 2. To get C3H8 on the product side of

the reaction, we need to reverse reaction 2.

3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) DH = +2220 kJ

C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) DH = -2220 kJ

Initial:

Final:


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Example (cont.)

Step 3: Add two new reactions together

to see what is left:

3C (gr) + 3O2 (g) 3CO2 (g) DH = -1182 kJ

3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) DH = +2220 kJ2

3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ


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Example (cont.)

Step 4: Compare previous reaction to final

reaction, and determine how to reach final

reaction:3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ

H2 (g) + 1/2O2 (g) H2O (l) DH = -286 kJ

3C (gr) + 4H2 (g) C3H8 (g)

Need to multiply second reaction by 4


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Example (cont.)

Step 4: Compare previous reaction to final

reaction, and determine how to reach final

reaction:3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ

4H2 (g) + 2O2 (g) 4H2O (l) DH = -1144 kJ

3C (gr) + 4H2 (g) C3H8 (g)


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Example (cont.)

Step 4 (cont.):

3C (gr) + 4H2O (l) C

3H

8(g) + 2O

2DH = +1038 kJ

4H2 (g) + 2O2 (g) 4H2O (l) DH = -1144 kJ

3C (gr) + 4H2 (g) C3H8 (g)D

H = -106 kJ


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process:



IfDH


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Another Example

Calculate DH for the following reaction:

H2(g) + Cl2(g) 2HCl(g)

Given the following:

NH3 (g) + HCl (g) NH4Cl(s) DH = -176 kJ

N2 (g) + 3H2 (g) 2NH3 (g) DH = -92 kJN2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) DH = -629 kJ


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Another Example (cont.)

Step 1: Only the first reaction contains the

product of interest (HCl). Therefore,

reverse the reaction and multiply by 2 to getstoichiometry correct.

NH3 (g) + HCl (g) NH4Cl(s) DH = -176 kJ

2NH4Cl(s) 2NH3 (g) + 2HCl (g) DH = 352 kJ


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Step 2. Need Cl2 as a reactant, therefore,

add reaction 3 to result from step 1 and see

what is left.2NH4Cl(s) 2NH3 (g) + 2HCl (g) DH = 352 kJ

N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) DH = -629 kJ

N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g)

DH = -277 kJ


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Step 3. Use remaining known reaction incombination with the result from Step 2 to

get final reaction.

N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) DH = -277 kJ

( N2 (g) + 3H2(g) 2NH3(g) DH = -92 kJ)

H2(g) + Cl2(g) 2HCl(g) DH = ?

Need to take middle reaction and reverse it


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Step 3. Use remaining known reaction incombination with the result from Step 2 to

get final reaction.

N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) DH = -277 kJ

2NH3(g) 3H2 (g) + N2 (g) DH = +92 kJ

H2(g) + Cl2(g) 2HCl(g) DH = -185 kJ

1


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In the above enthalpy diagram note that

H1 = H2 + H3

Hesss Law


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Enthalpies of Formation If 1 mol of compound is formed from its

constituent elements, then the enthalpy change for

the reaction is called the enthalpy of formation,

DHof.

Standard conditions (standard state): 1 atm and 25oC

Standard enthalpy, DHo, is the enthalpy measured

when everything is in its standard state.

Standard enthalpy of formation: 1 mol of

compound is formed from substances in their

standard states.


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Enthalpies of Formation Standard enthalpy of formation of the most stable

form of an element is zero.


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Hrxn =

H1 + H2 + H3

Enthalpies of FormationUsing Enthalpies of Formation to Calculate

Enthalpies of Reaction


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Enthalpies of FormationUsing Enthalpies of Formation to Calculate

Enthalpies of ReactionFor a reaction:

HOrxn = n HO

f(products) - n HO

f(reactants)


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process:



IfDH

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