CHAPTER 16

16.1

Every chemical reaction involves a change in energy (usually heat).

 When bonds are broken, energy (heat) needs to be added  When bonds are formed, energy (heat) is given off

What is the enthalpy (ΔH) of a reaction?

The enthalpy (ΔH) of a reaction is the energy change during the reaction. ΔH = H products - H reactants

The enthalpy (ΔH) of a reaction is the energy change during the reaction. ΔH = H products - H reactants An exothermic reaction occurs when there is more heat given off when the bonds are formed than is required to break the bonds.

The enthalpy (ΔH) of a reaction is the energy change during the reaction. ΔH = H products - H reactants An endothermic reaction occurs when there is more heat required to break the bonds than is given off when the bonds form.

ΔH is used to show the enthalpy of the reaction. +ΔH = ? -ΔH = ?

ΔH is used to show the enthalpy of the reaction. +ΔH = endothermic -ΔH = exothermic Where does the word energy appear in the equation of an exothermic reaction? ____?___ + A + B  AB + ___?____

An exothermic reaction graph actually looks like this. What is the activated complex? What is activation energy?

An endothermic reaction graph actually looks like this. What is the activated complex? What is activation energy?

How would a catalyst change this graph?

A catalyst lowers the __________, but does not affect the _________.

There are three ways to measure the change in enthalpy during a chemical reaction.

 Use a calorimeter and perform the reaction to collect data  Use a chart of molar enthalpies of formation  Use Hess’s Law

What is a calorimeter and how does it work?

Heat always moves from the substance with the higher temperature to the substance with the lower temperature, so if something loses heat, something gains heat.

When using a calorimeter, the heat lost/gained by the water is all based on the chemical reaction happening inside the “bomb”.

Heat always moves from the substance with the higher temperature to the substance with the lower temperature, so if something loses heat, something gains heat. When using a calorimeter, the heat lost/gained by the water is all based on the chemical reaction happening inside the “bomb”. q = heat

Heat always moves from the substance with the higher temperature to the substance with the lower temperature, so if something loses heat, something gains heat. When using a calorimeter, the heat lost/gained by the water is all based on the chemical reaction happening inside the “bomb”. q water = -q reaction

What is specific heat?

Specific heat is the capacity of a substance to absorb heat. P. 533

The heat transferred/absorbed has to take into account the mass of the substance, the temperature change, and the specific heat of the substance.

The formula for measuring this is q = C p m ΔT

The heat transferred/absorbed has to take into account the mass of the substance, the temperature change, and the specific heat of the substance. The formula for measuring this is q = C p m ΔT If an equal amount of heat were applied to carbon, water, and ammonia (all equal masses), which would have the highest temperature after the same amount of time? Lowest temperature? Why?

Example 1: How many joules are needed to raise the temperature of an iron nail that has a mass of 7.05g from 25ºC to 100ºC?

Example 2: How much will the temperature of 150g of liquid water change if it gains 40.0 joules?

Example 3: What is the specific heat of nickel if the temperature of a 32.3g sample of nickel is increased by 3.5ºC when 50J of heat is added?

Example 4: A 36.0g piece of lead at a temperature of 154.5ºC is added to a calorimeter containing 100.0g of water at 24.2ºC. The final temperature of the water is 25.6ºC. What is the specific heat of lead? (HINT – Remember q water = -q reaction )

Example 5: When a 28.7g sample of KI dissolves in 60.0g of water in a calorimeter, the temperature drops from 27.2ºC to 13.2ºC. Calculate the q of this process: KI(s)  K 1+ (aq) + I 1- (aq). What would the ∆H of KI(s)  K 1+ (aq) + I 1- (aq) be?

Example 6: A 1.0g sample of octane (C 8 H 18 ) is burned in a calorimeter containing 1200g of water. The temperature of the water rises from 25.00ºC to 33.20ºC. Calculate q of this process: C 8 H 18 (g) + 12½ O 2 (g)  8CO 2 (g) + 9H 2 O(g) What would the ∆H be?

Example 7: 2H 2 (g) + O 2 (g)  2H 2 O(l)ΔH = kJ What is the enthalpy change for the decomposition of 24.0g of H 2 O(l)?

Example 8: What is the enthalpy change if 27.1g of iodine reacts according to the equation below? H 2 (g) + I 2 (g)  2HI(g) ΔH = +26.5kJ

Example 9: How much heat will be transferred when 14.9g of ammonia reacts with excess O 2 according to the following equation? 4NH 3 + 5O 2  4NO + 6H 2 O ΔH = -1170kJ

Another way to calculate the energy change (ΔH ) is to use a chart where the molar enthalpy of formation is know for each compound. What is the molar enthalpy of formation?

The molar enthalpy of formation is the enthalpy change (ΔH) when one mole of a compound is formed from its elements at standard state (25ºC and 1 atm).

The molar enthalpy of formation is the enthalpy change (ΔH f ) when one mole of a compound is formed from its elements at standard state (25ºC and 1 atm). p. 862

The molar enthalpy of formation is the enthalpy change (ΔH f 0 ) when one mole of a compound is formed from its elements at standard state 0. (25ºC and 1 atm). p. 862 All elements have ΔH f = 0. A compound with a high –ΔH f is very stable. A compound with a +ΔH f or a small –ΔH f is unstable. (The more + the ΔH f the more unstable the compound.)

The molar enthalpy of formation is the enthalpy change (ΔH f 0 ) when one mole of a compound is formed from its elements at standard state 0. (25ºC and 1 atm). p. 862 When writing equations using ΔH f 0 the following rules apply:  Fractions can be used as coefficients.  States of matter are very important.  The ΔH is directly proportional to the number of moles.  The temperature at which the reaction takes place is relatively unimportant.

Example 10: Calculate the change in enthalpy for this reaction: H 2 (g) + CO 2 (g)  H 2 O(g) + CO(g)

Example 11: Calculate the change in enthalpy for this reaction: C 2 H 6 (g) + 7/2 O 2 (g)  2CO 2 (g) + 3H 2 O(g)

Example 12: Calculate the change in enthalpy when iron (III) oxide reacts with carbon to form iron and carbon dioxide gas.

Example 13: Write the equation when lead (IV) oxide is formed from its elements. What is the ∆H of this synthesis reaction?

Example 14: Write the equation when magnesium sulfate is formed from its elements. What is the ∆H of this synthesis reaction?

Example 14: Write the equation when lithium nitrate decomposes to form elements. What is the ∆H of this decomposition reaction?

What is Hess’s Law?

Hess’s Law can be used to predict the enthalpy changes in a chemical reaction based on either enthalpies of formation or enthalpy changes for each step of the reaction.C 2 H 6 + 7/2 O 2  2 CO 2 + 3H 2 O

1. C 2 H 6 + 7/2O 2  C 2 H 4 + H 2 + 7/2 O 2 2. C 2 H 4 + H 2 + 7/2O 2  2 CO H 2 O + H 2 + ½O CO H 2 O +H 2 + ½O 2  2 CO H 2 O T. C 2 H 6 + 7/2 O 2  2 CO H 2 O

Hess’s Law can be used to predict the enthalpy changes in a chemical reaction based on either enthalpies of formation or enthalpy changes for each step of the reaction.  If the reaction is reversed, the sign of ΔH changes.  If the equation is multiplied, the quantity of ΔH is multiplied.

Example 1: Calculate the standard enthalpy change, ΔH 0, for the formation of strontium carbonate (the material that gives the red color in fireworks) from its elements. Sr(s) + C(graphite) + 3/2 O 2 (g)  SrCO 3 (s) Sr(s) + ½ O 2 (g)  SrO(s)ΔH 0 = -592 kJ SrO(s) + CO 2 (g)  SrCO 3 (s)ΔH 0 = -234 kJ C(graphite) + O 2 (g)  CO 2 (g)ΔH 0 = -394 kJ

Example 2: The combination of coke and steam produces a mixture called coal gas, which can be used as a fuel or as a starting material for other reactions. If we assume coke can be represented by graphite, the equation for the production of coal gas is 2C(s) + 2H 2 O(g)  CH 4 (g) + CO 2 (g). Determine the standard enthalpy change for this reaction. C(s) + H 2 O(g)  CO(g) + H 2 (g)ΔH 0 = kJ CO(g) + H 2 O(g)  CO 2 (g) + H 2 (g)ΔH 0 = kJ CH 4 (g) + H 2 O(g)  3H 2 (g) + CO(g)ΔH 0 = kJ

Example 3: What is the ΔH 0 for burning ethanol, C 2 H 5 OH(l) to produce CO 2 (g) and H 2 O(g) C 2 H 5 OH(l) + 3O 2 (g)  2CO 2 (g) + 3H 2 O(l)ΔH 0 = kJ H 2 O(l)  H 2 O(g)ΔH 0 = 44 kJ

Example 4: What is the ΔH f 0 for H 2 S(g)? S(s) + O 2 (g)  SO 2 (g)ΔH 0 = -296 kJ H 2 (g) + 1/2O 2 (g)  H 2 O(l)ΔH 0 = -286 kJ H 2 S(g) + 3/2O 2 (g)  SO 2 (g) + H 2 O(l)ΔH 0 = -562 kJ

Example 5: What is the ∆H for the reaction below? N 2 H 4 (l) + CH 4 O(l)  CH 2 O(g) + N 2 (g) + 3H 2 (g) 2NH 3 (g)  N 2 H 4 (l) + H 2 (g) ∆H = +22.5kJ 2NH 3 (g)  N 2 (g) + 3H 2 (g) ∆H = +57.5kJ CH 2 O(g) + H 2 (g)  CH 4 O(l) ∆H = +81.2kJ

There are 3 ways the energy change of a chemical reaction can be determined:

 Use a calorimeter and perform the reaction to collect data  Use a chart of molar enthalpies of formation  Use Hess’s Law