the basis for calculating enthalpies of reaction is known as hess’s law: the overall enthalpy...

Hess’s Law

Calculating Enthalpies of Reactions

• The basis for calculating enthalpies of reaction is known as Hess’s law: the overall enthalpy change in a reaction is equal to the sum of enthalpy changes for the individual steps in the process.

• This means that the energy difference between reactants and products is independent of the route taken to get from one to the other.

If you know the reaction enthalpies of individual steps in an overall reaction, you can calculate the overall enthalpy without having to measure it experimentally.

Calculate the potential energy of each climber taking route 1 and route 2

-137KJ

+125KJ

+87KJ

-193KJ

+102KJ

-163KJ+52KJ

-147KJ

-269KJ

+7KJ

Regardless of the route the climber and the miner took they ended up having the same amount of

potential energy!!

5

Hess’s Law

StartFinish

A State Function: Path independent.

Both lines accomplished the same result, they went from start to finish. Net result = same.

• ∆Htarget = ∑∆Hknown

• ∆Htarget =∆Hrxn1+∆Hrxn2+∆Hrxn3+...

Hess’s Law Equation:

• For example: C + O2 CO2

occurs as 2 steps

C + ½O2 CO H = – 110.5 kJCO + ½O2 CO2 H = – 283.0 kJC + CO + O2 CO + CO2 H = – 393.5 kJ C + O2 CO2 H = – 393.5 kJ

Using Hess’s Law to find ΔH What is the enthalpy change for the

formation of two moles of nitrogen monoxide from its elements?

This reaction may be called the target equation to distinguish it clearly from other equationsN2(g) + O2(g) 2NO (g)

1. ½ N2(g) + O2(g) NO2(g) ΔH1θ = +34

kJ 2. NO(g) + ½ O2(g) NO2(g) ΔH2

θ = - 56 kJ

If we work with these two equations, which may be called known equations, and then add them together, we obtain the chemical equation for the formation of nitrogen monoxide

N2(g) + O2(g) 2NO (g)

If we look at the target eqn., it has 1 mole of N2 as reactant. How do we make it 1 mole?

2 x (½ N2(g) + O2(g) NO2(g) ΔH1θ = 2(+34 )kJ

2 x (NO2 (g) NO(g) + ½ O2(g) ΔH2θ = 2(+ 56 )kJ

N2(g) + 2O2(g) 2 NO2(g) ΔH1θ = 2(+34 )kJ

2 NO2 (g) 2NO(g) + O2(g) ΔH2θ = 2(+ 56 )kJ

N2(g) + 2O2(g) + 2NO2(g) 2NO2(g) + 2NO(g) + O2(g)

N2(g) + O2(g) 2NO(g)

∆Hθ = 2(+ 34) kJ + 2(+ 56) kJ

= + 68 kJ + 112 kJ

∆Hθ = + 180 kJ

• To demonstrate how to apply Hess’s law, we will work through the calculation of the enthalpy of formation for the formation of methane gas, CH4, from its elements,

hydrogen gas and solid carbon:

C(s) + 2H2(g) → CH4(g)fH 0 ?

• The component reactions in this case are the combustion reactions of carbon, hydrogen, and methane:

0 393.5 kJcH

0 285.8 kJcH

0 890.8 kJcH

H2(g) + ½O2(g) → H2O(l)

C(s) + O2(g) → CO2(g)

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)• The overall reaction involves the formation rather than the

combustion of methane, so the combustion equation for methane is reversed, and its enthalpy changed from negative to positive:

CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ∆H0 = +890.8 kJ

• Because 2 moles of water are used as a reactant in the above reaction, 2 moles of water will be needed as a product.

• Therefore, the coefficients for the formation of water reaction, as well as its enthalpy, need to be multiplied by 2:

2H2(g) + O2(g) → 2H2O(l) cH0 2( 285.8 kJ) • We are now ready to add the three equations together using

Hess’s law to give the enthalpy of formation for methane and the balanced equation.

0 393.5 kJcH 0 2( 285.8 kJ)cH

0 74.3 kJfH

2H2(g) + O2(g) → 2H2O(l)

C(s) + O2(g) → CO2(g)

C(s) + 2H2(g) → CH4(g)

CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ΔHθ = +890.8 kJ

• Using Hess’s law, any enthalpy of reaction may be calculated using enthalpies of formation for all the substances in the reaction of interest, without knowing anything else about how the reaction occurs.

• Mathematically, the overall equation for enthalpy change will be in the form of the following equation:

∆H0 = sum of [( of products) × (mol of products)] – sum of [( of reactants) × (mol of reactants)]

fH 0

fH 0

• Hess’s law allows us to add equations.

• We add all reactants, products, & H values.

• We can also show how these steps add together via an “enthalpy diagram” …

Exothermic Enthalpy Diagram

Endothermic Enthalpy Diagram

Steps in drawing enthalpy diagrams

1. Balance the equation(s).

2. Sketch a rough draft based on H values.

3. Draw the overall chemical reaction as an enthalpy diagram (with the reactants on one line, and the products on the other line).

4. Draw a reaction representing the intermediate step by placing the relevant reactants on a line.

5. Check arrows: Start: two leading away

Finish: two pointing to finishIntermediate: one to, one away

6. Look at equations to help complete balancing (all levels must have the same # of all atoms).

7. Add axes and H values.

C + O2 CO2 H = – 393.5 kJ

Reactants

Intermediate

Products

C + O2

CO2

CO

Ent

halp

y

Note: states such as (s) and (g) have been ignored to reduce clutter on these slides. You should include these in your work.

H = – 110.5 kJ

H = – 283.0 kJ

H = – 393.5 kJ

+ ½ O2

C + ½ O2 CO H = – 110.5 kJCO + ½ O2 CO2 H = – 283.0 kJ

How much energy is lost in the formation of C2H5OH(l) ?

C2H4(g) + H2O(l) C2H5OH(l)

Given:• C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)

H= –1411.1 kJ

• 2CO2(g) + 3H2O(l) C2H5OH(l) + 3O2(g)

• H= +1367.1 kJ

C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H= –1411.1 kJ2CO2(g) + 3H2O(l) C2H5OH(l) + 3O2(g) H= +1367.1

kJC2H4(g) + H2O(l) C2H5OH(l)

Reactants Products

Intermediate

C2H4(g) + H2O(l)C2H5OH(l)

2CO2(g) + 3H2O(l)

Ent

halp

yH

=

–

1411

.1 k

J

H = +1367.1 kJ

H= – 44.0 kJ

+ 3O2(g)+ 3O2(g)

H= – 44.0 kJ

Solution:

Two Rules to Follow:

1. If a chemical equation is reversed, then the sign of ∆H changes

2. If the coefficients of a chemical equation are altered by multiplying or dividing by a constant factor, then the value of ∆H is altered in the same way

We may need to manipulate equations further: 2Fe + 1.5O2 Fe2O3 H=?,

Given:

Fe2O3 + 3CO 2Fe + 3CO2 H= – 26.74 kJ CO + ½ O2 CO2 H= – 282.96 kJ

1: Align equations based on reactants/products.2: Multiply based on final reaction.3: Add equations.

2Fe + 1.5O2 Fe2O3

3CO + 1.5 O2 3CO2 H= – 848.88 kJ

2Fe + 3CO2 Fe2O3 + 3CO H= + 26.74 kJ CO + ½ O2 CO2 H= – 282.96 kJ

H= – 822.14 kJ

Flip equation =Flip sign

Multiply coefficients and values

Sample Problem BCalculate the enthalpy of reaction for the combustion of nitrogen monoxide gas, NO, to form nitrogen dioxide gas, NO2, as given in the following equation.

NO(g) + ½O2(g) → NO2(g)

Use the enthalpy-of-formation data. Solve by combining the known thermochemical equations. Given: ½ N2(g) + ½ O2(g) NO(g) ∆Hf

θ +90.29 kJ

½ N2(g) + O2(g) NO2(g) ∆Hfθ +33.2 kJ

Unknown: ∆Hθ for NO(g) + ½ O2(g) NO2(g)

Solution:

Using Hess’s law, combine the given thermochemical equations in such a way as to obtain the unknown equation, and its ∆H0 value.

The desired equation is:g + g g1

2 22NO( ) O ( ) NO ( )

g g + g = k1 122 2 f2

0NO( ) N ( ) O ( H – 90.29) J

g g g k012 2 2 f2 N ( ) + O ( ) NO ( ) ΔH =+33.2 J

The other equation should have NO2 as a product, so we can use the second given equation as is:

Reversing the first given reaction and its sign yields the following thermochemical equation:

We can now add the equations and their ∆H0 values to obtain the unknown ∆H0 value.

g g g k012 2 2 f2 N ( ) + O ( ) NO ( ) H =+33.2 J

g g + g = k1 122 2 f2

0NO( ) N ( ) O ( H – 90.29) J

0 57.1 kJH g + g g12 22NO( ) O ( ) NO ( )

Determining Enthalpy of Formation

• When carbon is burned in a limited supply of oxygen, carbon monoxide is produced:

s + g g122C( ) O ( ) CO( )

• The above overall reaction consists of two reactions:

1) carbon is oxidized to carbon dioxide

2) carbon dioxide is reduced to give carbon monoxide.

• Because these two reactions occur simultaneously, it is not possible to directly measure the enthalpy of formation of CO(g) from C(s) and O2(g).

• We do know the enthalpy of formation of carbon dioxide and the enthalpy of combustion of carbon monoxide:

fH 02 2C(s) + O (g) CO (g) 393.5 kJ/mol

cg g g H 012 22CO( ) + O ( ) CO ( ) 283.0 kJ/mol

g g g H 012 22 CO ( ) CO( ) + O ( ) 283.0 kJ/mol

H02 2C(s) + O (g) CO (g) 393.5 kJ/mol

• We reverse the second equation because we need CO as a product. Adding gives the desired enthalpy of formation of carbon monoxide.

0 110.5 kJH s + g g122C( ) O ( ) CO( )

Determine the heat of reaction for the reaction:

Target 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) H= ?

Using the following sets of reactions:

(1) N2(g) + O2(g) 2NO(g) H = 180.6 kJ

(2) N2(g) + 3H2(g) 2NH3(g) H = -91.8 kJ

(3) 2H2(g) + O2(g) 2H2O(g) H = -483.7 kJ

Hint: The three reactions must be algebraically manipulated to sum up to the desired reaction.and.. the H values must be treated accordingly.

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

Using the following sets of reactions:

(1) N2(g) + O2(g) 2NO(g) H = 180.6 kJ

(2) N2(g) + 3H2(g) 2NH3(g) H = -91.8 kJ

(3) 2H2(g) + O2(g) 2H2O(g) H = -483.7 kJ

Goal:

NH3:O2 :

NO:

H2O:

(2)(Reverse and x 2) 4NH3 2N2 + 6H2 H = +183.6 kJ Found in more than one place, SKIP IT (its hard).

(1) (Same x2) 2N2 + 2O2 4NO H = 361.2 kJ

(3)(Same x3) 6H2 + 3O2 6H2O H = -1451.1 kJ

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)Goal:

NH3:O2 :

NO:

H2O:

Reverse and x2 4NH3 2N2 + 6H2 H = +183.6 kJ Found in more than one place, SKIP IT.

x2 2N2 + 2O2 4NO H = 361.2 kJ

x3 6H2 + 3O2 6H2O H = -1451.1 kJ

Cancel terms and take sum.

4NH3+ 5O2 4NO + 6H2O H = -906.3 kJ

Is the reaction endothermic or exothermic?

H = 183.6 kJ + 361.2 kJ + (-1451kJ)

Determine the heat of reaction for the reaction:

TARGET C2H4(g) + H2(g) C2H6(g) H = ?

Use the following reactions:

(1) C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H = -1401 kJ

(2) C2H6(g) + 7/2O2(g) 2CO2(g) + 3H2O(l) H = -1550 kJ

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

Consult your neighbour if necessary.

Determine the heat of reaction for the reaction:

Goal: C2H4(g) + H2(g) C2H6(g) H = ?

Use the following reactions:

(1) C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H = -1401 kJ

(2) C2H6(g) + 7/2O2(g) 2CO2(g) + 3H2O(l) H = -1550 kJ

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

C2H4(g) :use 1 as is C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H = -1401 kJ H2(g) :# 3 as is H2(g) + 1/2O2(g) H2O(l) H = -286 kJC2H6(g) : rev #2 2CO2(g) + 3H2O(l) C2H6(g) + 7/2O2(g) H = +1550 kJ

C2H4(g) + H2(g) C2H6(g) H = -137 kJ

Hess’s Law Example Problem

Calculate H for the combustion of methane, CH4 using the equations aboveAns = -890.36 kJ

    Reaction Hfo  

C + 2H2 CH4 -74.80 kJ

C + O2 CO2 -393.50 kJ

H2 + ½ O2 H2O -285.83 kJ