Hess's Law, Exercises of Chemistry

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Hess’s Law

Start

Finish

Path independent

Both lines accomplished the same result,

they went from start to finish.

Net result = same.

C + O

2

 CO

2

H = – 393.

kJ

Reactants

Intermediate

Products

C + O

2

CO

2

CO

Enthalpy

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 O

2

C + ½ O

2

 CO H = – 110.5 kJ

CO + ½ O

2

 CO

2

H = – 283.0 kJ

Determine the heat of reaction for the reaction:

4NH 3

(g) + 5O 2

(g)  4NO(g) + 6H 2

O(g)

Using the following sets of reactions:

(1) N 2

(g) + O 2

(g)  2NO(g)  H = 180.6 kJ

(2) N 2

(g) + 3H 2

(g)  2NH 3

(g)  H = - 91.8 kJ

(3) 2H 2

(g) + O 2

(g)  2H 2

O(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.

4NH 3

(g) + 5O 2

(g)  4NO(g) + 6H 2

Goal: O(g)

NH 3

:

O 2

:

NO:

H 2

O:

Reverse and x2 (^4) NH 3

 2 N 2

+ 6 H 2

 H = +183.6 kJ

Found in more than one place, SKIP IT.

x2 (^2) N 2

+ 2 O 2

 4 NO  H = 361.2 kJ

x3 (^6) H 2

+ 3 O 2

 6 H 2

O  H = - 1451.1 kJ

Cancel terms and take sum.

4 NH 3

+ 5 O

2

 4 NO +^6 H

2

O  H = - 906.3 kJ

Is the reaction endothermic or exothermic?

Calculate the standard enthalpy of formation of CS 2

( l ) given

that:

C(graphite) + O 2

( g ) CO 2

( g ) H

0 = -393.5 kJ rxn

S(rhombic) + O 2

( g ) SO 2

( g ) H

0 = -296.1 kJ rxn

CS 2 ( l ) + 3O 2 ( g ) CO 2 ( g ) + 2SO 2 ( g ) H

0 = -1072 kJ rxn

1. Write the enthalpy of formation reaction for CS 2

C(graphite) + 2S(rhombic) CS 2

( l )

2. Add the given rxns so that the result is the desired rxn.

rxn

C(graphite) + O 2 ( g ) CO 2 ( g ) H

0 = -393.5 kJ

2S(rhombic) + 2O 2 ( g ) 2SO 2 ( g ) H

0 = -296.1x2 kJ rxn

CO 2

( g ) + 2SO 2

( g ) CS 2

( l ) + 3O 2

( g ) H

0 = +1072 kJ rxn

C(graphite) + 2S(rhombic) CS 2

( l )

H

0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ 6.5 rxn

Determine the heat of reaction for the reaction:

TARGET  C 2

H 4

(g) + H 2

(g)  C 2

H 6

(g)

Use the following reactions:

(1) C 2

H 4

(g) + 3O 2

(g)  2CO 2

(g) + 2H 2

O(l)  H = -1401 kJ

(2) C 2

H 6

(g) + 7/2O 2

(g)  2CO 2

(g) + 3H 2

O(l)  H = -1550 kJ

(3) H 2

(g) + 1/2O 2

(g)  H 2

O(l)  H = -286 kJ

Consult your neighbor if necessary.

Benzene (C 6

H 6

) burns in air to produce carbon dioxide and liquid

water. How much heat is released per mole of benzene

combusted? The standard enthalpy of formation of benzene is

49.04 kJ/mol.

2C 6

H 6

( l ) + 15O 2

( g ) 12CO 2

( g ) + 6H 2

O ( l )

H

0

rxn

n H

0 (products) f

= ^ n H

0 (reactants) f

• 

H

0

rxn

6 H

0 (H 2

O) f

12 H

0 (CO 2

) f

= [^ +^ ]^ -^2 H

0 (C 6

H 6

) f

[ ]

H

0

rxn

= [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6534.88 kJ

-6534.88 kJ

2 mol

= - 3267.44 kJ/mol C 6

H 6

Hess’s Law Example ProblemHess’s Law Example Problem

Calculate  H for the combustion of methane, CH 4

CH 4

+ 2O 2

 CO 2

+ 2H 2

O

Reaction  H

o

C + 2H 2

 CH 4 -74.80 kJ

C + O 2

 CO 2 -393.50 kJ

H 2

+ ½ O 2

 H 2

O (^) -285.83 kJ

Summary:

Used for reactions that

cannot be determined

experimentally

Summary:

 H is independent of

the path taken

Enthalpy ValuesEnthalpy Values

H H 22

(g) + 1/2 O(g) + 1/2 O (^22)

(g) --> H(g) --> H 22

O(g)O(g)

∆ ∆H˚ = -242 kJH˚ = -242 kJ

2 H 2 H 22

(g) + O(g) + O 22

(g) --> 2 H(g) --> 2 H 22

O(g)O(g)

∆ ∆H˚ = -484 kJH˚ = -484 kJ

H H 22

O(g) ---> HO(g) ---> H 22

(g) + 1/2 O(g) + 1/2 O (^22)

(g)(g)

∆ ∆H˚ = +242 kJH˚ = +242 kJ

H H 22

(g) + 1/2 O(g) + 1/2 O 22

(g) --> H(g) --> H 22

O(liquid)O(liquid)

∆ ∆H˚ = -286 kJH˚ = -286 kJ

Depend on how the reaction is written and on Depend on how the reaction is written and on

phases of reactants and products phases of reactants and products

Estimating Heats of Reaction

Using BOND ENERGIES

• (^) Chemical bonding is actually a continuum, from very ionic

(i.e., complete electron transfer) to pure covalent (i.e., equal

electron sharing):

• (^) Very ionic  ----------------------  Pure Covalent

Incr. bond decreasing bond

"strength“  ---------------  "strength“

• (^) Bond "strength" can thus be thought of in terms of the extent of

electron transfer.

• (^) Another way to think about bond "strength" is how much

energy is needed to break a bond apart (or how much energy is

released when a bond forms).