TYPES OF HEAT REACTIONS ASSIGNMENT

Answer:

ΔH = −153 kJ

2. Another Example of the Measurement of an Enthalpy Change
A gummy bear contains 2.67 g sucrose, C₁₂H₂₂O₁₁. When it reacts with 7.19 g potassium chlorate, KClO₃, 43.7 kJ of heat are produced. Determine the enthalpy change for the reaction

.

Solution
We have  available, and  available. Since  is needed, C₁₂H₂₂O₁₁ is the excess reactant and KClO₃ is the limiting reactant.

The reaction uses 8 mol KClO₃, and the conversion factor is , so we have . The enthalpy change for this reaction is −5960 kJ, and the thermochemical equation is:

Check Your Learning
When 1.42 g of iron reacts with 1.80 g of chlorine, 3.22 g of FeCl₂(s) and 8.60 kJ of heat is produced. What is the enthalpy change for the reaction when 1 mole of FeCl₂(s) is produced?

Answer:

ΔH = −338 kJ

3.

Using Enthalpy of Combustion
As Figure 3 suggests, the combustion of gasoline is a highly exothermic process. Let us determine the approximate amount of heat produced by burning 1.00 L of gasoline, assuming the enthalpy of combustion of gasoline is the same as that of isooctane, a common component of gasoline. The density of isooctane is 0.692 g/mL.

Solution
Starting with a known amount (1.00 L of isooctane), we can perform conversions between units until we arrive at the desired amount of heat or energy. The enthalpy of combustion of isooctane provides one of the necessary conversions. Table 2 gives this value as −5460 kJ per 1 mole of isooctane (C₈H₁₈).

Using these data,

The combustion of 1.00 L of isooctane produces 33,100 kJ of heat. (This amount of energy is enough to melt 99.2 kg, or about 218 lbs, of ice.)

Note: If you do this calculation one step at a time, you would find:

Check Your Learning
How much heat is produced by the combustion of 125 g of acetylene?

Answer:

6.25 × 10³ kJ

Evaluating an Enthalpy of Formation
Ozone, O₃(g), forms from oxygen, O₂(g), by an endothermic process. Ultraviolet radiation is the source of the energy that drives this reaction in the upper atmosphere. Assuming that both the reactants and products of the reaction are in their standard states, determine the standard enthalpy of formation,  of ozone from the following information:

Solution
 is the enthalpy change for the formation of one mole of a substance in its standard state from the elements in their standard states. Thus,  for O₃(g) is the enthalpy change for the reaction:

For the formation of 2 mol of O₃(g), . This ratio, , can be used as a conversion factor to find the heat produced when 1 mole of O₃(g) is formed, which is the enthalpy of formation for O₃(g):

Therefore, .

Check Your Learning
Hydrogen gas, H₂, reacts explosively with gaseous chlorine, Cl₂, to form hydrogen chloride, HCl(g). What is the enthalpy change for the reaction of 1 mole of H₂(g) with 1 mole of Cl₂(g) if both the reactants and products are at standard state conditions? The standard enthalpy of formation of HCl(g) is −92.3 kJ/mol.

For the reaction 

Writing Reaction Equations for 
Write the heat of formation reaction equations for:

(a) C₂H₅OH(l)

(b) Ca₃(PO₄)₂(s)

Solution
Remembering that  reaction equations are for forming 1 mole of the compound from its constituent elements under standard conditions, we have:

(a) 

(b) 

Note: The standard state of carbon is graphite, and phosphorus exists as P₄.

Check Your Learning
Write the heat of formation reaction equations for:

(a) C₂H₅OC₂H₅(l)

(b) Na₂CO₃(s)

Answer:

(a) ;

(b) 

6. Stepwise Calculation of  Using Hess’s Law
Determine the enthalpy of formation,  , of FeCl₃(s) from the enthalpy changes of the following two-step process that occurs under standard state conditions:

Solution
We are trying to find the standard enthalpy of formation of FeCl₃(s), which is equal to ΔH° for the reaction:

Looking at the reactions, we see that the reaction for which we want to find ΔH° is the sum of the two reactions with known ΔH values, so we must sum their ΔHs:

The enthalpy of formation, , of FeCl₃(s) is −399.5 kJ/mol.

Check Your Learning
Calculate ΔH for the process:

from the following information:

Answer:

66.4 kJ

A More Challenging Problem Using Hess’s Law
Chlorine monofluoride can react with fluorine to form chlorine trifluoride:

(i) 

Use the reactions here to determine the ΔH° for reaction (i):

(ii) 

(iii) 

(iv) 

Solution
Our goal is to manipulate and combine reactions (ii), (iii), and (iv) such that they add up to reaction (i). Going from left to right in (i), we first see that ClF(g) is needed as a reactant. This can be obtained by multiplying reaction (iii) by , which means that the ΔH° change is also multiplied by :

Next, we see that F₂ is also needed as a reactant. To get this, reverse and halve reaction (ii), which means that the ΔH° changes sign and is halved:

To get ClF₃ as a product, reverse (iv), changing the sign of ΔH°:

Now check to make sure that these reactions add up to the reaction we want:

Reactants  and  cancel out product O₂; product  cancels reactant ; and reactant  is cancelled by products  and OF₂. This leaves only reactants ClF(g) and F₂(g) and product ClF₃(g), which are what we want. Since summing these three modified reactions yields the reaction of interest, summing the three modified ΔH° values will give the desired ΔH°:

Check Your Learning
Aluminum chloride can be formed from its elements:

(i) 

Use the reactions here to determine the ΔH° for reaction (i):

(ii) 

(iii) 

(iv) 

(v) 

Answer:

−1407 kJ

Using Hess’s Law
What is the standard enthalpy change for the reaction:

Solution: Using the Equation
Use the special form of Hess’s law given previously:

Solution: Supporting Why the General Equation Is Valid
Alternatively, we can write this reaction as the sum of the decompositions of 3NO₂(g) and 1H₂O(l) into their constituent elements, and the formation of 2HNO₃(aq) and 1NO(g) from their constituent elements. Writing out these reactions, and noting their relationships to the  values for these compounds , we have:

Summing these reaction equations gives the reaction we are interested in:

Summing their enthalpy changes gives the value we want to determine:

So the standard enthalpy change for this reaction is ΔH° = −138.4 kJ.

Note that this result was obtained by (1) multiplying the  of each product by its stoichiometric coefficient and summing those values, (2) multiplying the  of each reactant by its stoichiometric coefficient and summing those values, and then (3) subtracting the result found in (2) from the result found in (1). This is also the procedure in using the general equation, as shown.

Check Your Learning
Calculate the heat of combustion of 1 mole of ethanol, C₂H₅OH(l), when H₂O(l) and CO₂(g) are formed. Use the following enthalpies of formation: C₂H₅OH(l), −278 kJ/mol; H₂O(l), −286 kJ/mol; and CO₂(g), −394 kJ/mol.

Answer:

−1368 kJ/mol