Learning Objectives Use the qualitative or quantitative heat transfer information in a chemical equation to classify a reaction as endothermic or exothermic. Use the endothermic or exothermic classification of a reaction to incorporate qualitative or quantitative heat transfer information into a chemical equation. Apply a stoichiometric conversion factor to convert between the amount of heat that is transferred during a chemical reaction and the molar quantity of a substance that participates in that reaction. Apply multiple conversion factors to solve problems that involve complex molar relationships. The previous section of this chapter discussed the energetic impacts of the molecular-level transformations that occur during chemical changes. In order for a chemical change to occur, bonds that are present in the reactant molecules must be broken, so that the associated atoms can separate and rearrange, and new bonds are ultimately created. As a result, the overall combination of bonds and, therefore, the total amounts of potential energy that are present before and after a chemical reaction are not identical. The Law of Conservation of Energy mandates that this difference in energy must be offset by a corresponding heat transfer between the reaction system and its surroundings. If the combined energy of the reactants is greater than the total energy of the products, the excess energy is released as heat. These reactions, in which heat is generated as a product, are classified as exothermic. In an endothermic reaction, the reactants have less energy, overall, than the products, and heat must be absorbed as a reactant to initiate the corresponding chemical change. This heat transfer information can be incorporated into chemical equations, using both qualitative symbolism and quantitative numerical values. Qualitatively Symbolizing Heat Transfers in Chemical Equations The variable for heat, q, can be written on the left or right side of a reaction arrow to indicate that heat is absorbed as a reactant or released as a product, respectively. For example, consider the decomposition of nitroglycerin, C 3 H 5 N 3 O 9 , which is classified as a "high explosive" because of its tendency to spontaneously and violently explode if disturbed by a minor impact or frictional force. Upon detonation, this molecule reacts to form molecular nitrogen, N 2 , carbon dioxide, CO 2 , water, H 2 O, and molecular oxygen, O 2 , which are all highly-stable covalent molecules. Because the variable for heat, q, is written on the right side of the reaction arrow in the chemical equation that is shown below, heat is generated as a product, and, therefore, the corresponding reaction is classified as exothermic. \(4 \; \ce{C_3H_5N_3O_9} \left( l \right) \rightarrow 6 \; \ce{N_2} \left( g \right) + 12 \; \ce{CO_2} \left( g \right) + 10 \; \ce{H_2O} \left( g \right) + \ce{O_2} \left( g \right) + \text{q} \) However, as shown in Example \(\PageIndex{1}\), not all decomposition reactions are exothermic. In general, there is no correlation between the classification of a reaction and its corresponding heat transfer. Therefore, in order to classify a chemical change as endothermic or exothermic, the reaction must be studied in a laboratory setting. Example \(\PageIndex{1}\) The decomposition reaction that is represented by the following chemical equation is endothermic. Incorporate the variable for heat into this equation, accordingly. \(\ce{CaCO_3} \left( s \right) \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right)\) Solution In an endothermic reaction, heat must be absorbed into the reaction system. Therefore, the variable for heat, q, must be written on the reactant , or left , side of the reaction arrow, as shown below. \(\ce{CaCO_3} \left( s \right) + \text{q} \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right)\) Exercise \(\PageIndex{1}\) The combination reaction that occurs when elemental sodium interacts with molecular chlorine is exothermic. Write a balanced chemical equation that represents this chemical change, and incorporate the variable for heat into the equation, accordingly. (States of matter are not required.) Answer Based on the given information, two reactants, elemental sodium, Na, and molecular chlorine, Cl 2 , will participate in this reaction. The chemical formula for molecular chlorine, Cl 2 , is derived by applying the Chapter 3 rules for determining diatomic covalent chemical formulas. In a combination reaction, the components of two or more reactants bond with one another to form a single product . In this particular reaction, the product molecule must contain both sodium, Na, and chlorine, Cl. However, the subscripts that are present within a chemical formula are solely dependent on the elemental, ionic, or covalent nature of the corresponding substance. Therefore, the chemical formula of the compound that is expected to form in this reaction cannot be obtained simply by rewriting the above-mentioned chemical information as a single formula. Because sodium, Na , is a metal, and chlorine, Cl, is a non-metal, the formula of the compound that contains these elements can only be established by applying the Chapter 3 rules for determining ionic chemical formulas. In this reaction, the sodium ion, Na +1 , bonds with the chloride ion, Cl –1 , to produce sodium chloride, NaCl . Writing each of these chemical formulas on the appropriate side of a reaction arrow results in the unbalanced chemical equation that is shown below. ___ \(\ce{Na}\) + ___ \( \ce{Cl_2} \rightarrow \) ___ \(\ce{NaCl}\) In order to balance a chemical equation, the quantities of each type of element and polyatomic ion that are present in the reactants and the products of the reaction must be determined. No polyatomic ions are present in any of the formulas that are shown above. Therefore, only individual elements must be considered for this reaction. The quantities of sodium, Na, and chlorine, Cl, that are present in the given equation are summarized in the table that is shown below. Element or Ion Reactants Products Balanced Na 1 1 NaN Cl 2 1 NaN Since both sides of the equation contain equal amounts of sodium, Na, that element is balanced. However, chlorine, Cl, is not balanced, because it is present in different quantities on the reactant and product sides of the reaction arrow. Therefore, one or more coefficients must be written in the "blanks" above, in order to balance this reaction. In order to balance chlorine, Cl, a coefficient should be written in the "blank" that corresponds to sodium chloride, NaCl, on the right side of the equation, as fewer chlorines are present on this side of the reaction arrow. The value of this coefficient, 2 , is determined by dividing the larger quantity of this element, 2, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{Na}\) + ___ \( \ce{Cl_2} \rightarrow \) 2 \(\ce{NaCl}\) As sodium chloride, NaCl, contains both sodium, Na, and chlorine, Cl, incorporating this coefficient alters the amounts of both of these elements on the product side of the equation. The updated quantities of these elements are reflected in the table that is shown below. Inserting this coefficient balances chlorine, as intended. However, the quantities in which sodium, Na, are present, which had been identical, are now unequal. Therefore, this element must now be balanced, as well, and one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced Na 1 \( \cancel{\rm{1} } \) 2 NaN Cl 2 \( \cancel{\rm{1} } \) 2 NaN In order to balance sodium, Na, a coefficient should be written in the "blank" that corresponds to this element on the left side of the equation, as fewer sodiums are present on this side of the reaction arrow. The value of this coefficient, 2 , is determined by dividing the larger quantity of this element, 2, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. 2 \(\ce{Na}\) + ___ \( \ce{Cl_2} \rightarrow \) 2 \(\ce{NaCl}\) As sodium, Na, is present in its atomic form, incorporating this coefficient only alters the amount of this element on the reactant side of the equation, as reflected in the table that is shown below. Inserting this coefficient balances sodium, Na, as intended. Therefore, all of the components of this equation are now balanced. Element or Ion Reactants Products Balanced Na \( \bcancel{\rm{1} } \) 2 \( \cancel{\rm{1} } \) 2 NaN Cl 2 \( \cancel{\rm{1} } \) 2 NaN No fractional coefficients are written in the equation that is shown above. Furthermore, because of the unwritten "1" that is understood to occupy the second "blank" in this equation, these coefficients cannot be divided. Finally, the heat transfer that occurs in the corresponding chemical reaction must be indicated by incorporating the variable for heat, q, into the reaction equation. Since heat is released as a product in an exothermic reaction, the variable for heat, q, must be written on the right side of the reaction arrow, as shown below. 2 \(\ce{Na}\) + \( \ce{Cl_2} \rightarrow \) 2 \(\ce{NaCl}\) + \(\text{q}\) Quantifying Heat Transfers in Chemical Reactions The energy values that are associated with most types of bonds are known quantities that can be utilized to calculate the total energies of the reactants and the products of a chemical reaction. The difference between these sums corresponds to the amount of heat energy that is absorbed or released in a particular chemical reaction. The resultant numerical value, which is normally expressed in kilojoules (kJ), can be directly incorporated as a quantitative reactant or product in an endothermic or exothermic reaction, respectively. For example, the decomposition of nitroglycerin, C 3 H 5 N 3 O 9 , produces 5,660 kilojoules of heat energy. Therefore, instead of incorporating the variable for heat, q, into the chemical equation for this reaction, the quantitative amount of heat that is generated can be written on the right side of the reaction arrow, as shown below. \(4 \; \ce{C_3H_5N_3O_9} \left( l \right) \rightarrow 6 \; \ce{N_2} \left( g \right) + 12 \; \ce{CO_2} \left( g \right) + 10 \; \ce{H_2O} \left( g \right) + \ce{O_2} \left( g \right) + 5,660 \; \text{kJ} \) Furthermore, because this chemical reaction is balanced, stoichiometric equalities can be developed to relate the quantity of heat that is transferred during the reaction to the molar quantities of the chemicals that are involved in the reaction. For example, calculate how many moles of nitroglycerin, C 3 H 5 N 3 O 9 , must be consumed to produce 87,200 kilojoules of heat. Because the chemical that is referenced in the problem, nitroglycerin, C 3 H 5 N 3 O 9 , and an energetic quantity are both present in the reaction equation that is shown above, a stoichiometric equality should be developed and applied to solve this problem. The left side of this equality should be created using the equality pattern that was presented and discussed in Section 4.26. Since stoichiometry studies the molar relationships that exist between chemicals and quantities that are involved in chemical reactions, the primary unit for the chemical portion of the equality is "mol," and the chemical formula of nitroglycerin, C 3 H 5 N 3 O 9 , is incorporated as the secondary unit on the left side of the equality that is being developed. Finally, since the coefficient that is associated with with nitroglycerin, C 3 H 5 N 3 O 9 , is a " 4 " in the corresponding chemical equation, a 4 is inserted into the numerical position on the left side of this stoichiometric equality. Because this problem references an energetic quantity, rather than a second chemical that participates in the reaction, the amount of heat that is generated in the reaction is written on the right side of this stoichiometric equality, as shown below. 4 mol C 3 H 5 N 3 O 9 = 5,660 kJ The information that is contained in this equality can be re-written in the form of a conversion factor, which can be applied to bring about the desired unit transformation, as shown below. \( {87,200 \; \cancel{\rm{kJ}}} \times\) \( \dfrac{4 \; \rm{mol} \; \rm{C_3H_5N_3O_9}}{5,660 \; \cancel{\rm{kJ} }}\) = \( {\text {61.62544...}}\) \({\rm{mol} \; \rm{C_3H_5N_3O_9}}\) ≈ \( {\text {61.6}}\) \({\rm{mol} \; \rm{C_3H_5N_3O_9}}\) Exercise \(\PageIndex{2}\) Consider the following chemical equation. ___ \(\ce{C_2H_2} \left( g \right) + \) ___ \( \ce{O_2} \left( g \right) \rightarrow\) ___ \( \ce{CO_2} \left( g \right) + \) ___ \( \ce{H_2O} \left( g \right) + 2,510 \; \text{kJ} \) Balance this equation by writing coefficients in the "blanks," as necessary. Classify this reaction as a combination, a decomposition, a single replacement, a double replacement, or a combustion reaction. Classify this reaction as endothermic or exothermic. Calculate how many kilojoules of heat are transferred if 3.16 x 10 24 molecules of water are generated in this reaction. Answer a Coefficients are incorporated into a chemical equation in order to account for any relative differences between the chemical formulas of the reactants and products that are involved in the corresponding chemical reaction. However, the numerical quantity that is present in the reaction, 2,510 kJ, is not a chemical material and, therefore, does not contain elements or polyatomic ions. As a result, there is no need to associate a balancing coefficient with the heat energy that is incorporated into this equation. In order to balance a chemical equation, the quantities of each type of element and polyatomic ion that are present in the reactants and the products of the reaction must be determined. No polyatomic ions are present in any of the formulas that are shown above. Therefore, only individual elements must be considered for this reaction. The quantities of carbon, C, hydrogen, H, and oxygen, O, that are present in the given equation are summarized in the table that is shown below. Note that oxygen is present in both of the chemical products that are generated during this reaction. In order to determine the number of oxygens that are present on the product side of this equation, the subscripts on both oxygens must be added . Element or Ion Reactants Products Balanced C 2 1 NaN H 2 2 NaN O 2 3 NaN Since both sides of the equation contain equal amounts of hydrogen, H, that element is balanced. Neither carbon, C, nor oxygen, O, is balanced, because each of these elements is present in different quantities on the reactant and product sides of the reaction arrow. Therefore, one or more coefficients must be written in the "blanks" above, in order to balance this reaction. As stated above, oxygen is present in both of the chemical products that are generated during this reaction and, therefore, should not be the first element to be balanced in this equation. In order to balance carbon, C, a coefficient should be written in the "blank" that corresponds to carbon dioxide, CO 2 , on the right side of the equation, as fewer carbons are present on this side of the reaction arrow. The value of this coefficient, 2 , is determined by dividing the larger quantity of this element, 2, by its smaller count, 1. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{C_2H_2} \left( g \right) + \) ___ \( \ce{O_2} \left( g \right) \rightarrow\) 2 \( \ce{CO_2} \left( g \right) + \) ___ \( \ce{H_2O} \left( g \right) + 2,510 \; \text{kJ} \) As carbon dioxide, CO 2 , contains both carbon, C, and oxygen, O, incorporating this coefficient alters the amounts of both of these elements on the product side of the equation. The updated quantities of these elements are reflected in the table that is shown below. As stated above, because oxygen is present in both of the chemical products that are generated during this reaction, the quantities of oxygen in both of these products must be added to determine the amount of oxygen that is present on the product side of this equation. Inserting this coefficient balances carbon, C, as intended. The quantities in oxygen, O, are present are still unequal. Therefore, one or more additional coefficients must be written in the "blanks" above, in order to balance this reaction. Element or Ion Reactants Products Balanced C 2 \( \cancel{\rm{1} } \) 2 NaN H 2 2 NaN O 2 \( \cancel{\rm{3} } \) 5 NaN In order to balance oxygen, O, a coefficient should be written in the "blank" that corresponds to molecular oxygen, O 2 , on the left side of the equation, as fewer oxygens are present on this side of the reaction arrow. The value of this coefficient, 2.5 , is determined by dividing the larger quantity of this element, 5, by its smaller count, 2. Inserting this coefficient results in the chemical equation that is shown below. ___ \(\ce{C_2H_2} \left( g \right) + \) 2.5 \( \ce{O_2} \left( g \right) \rightarrow\) 2 \( \ce{CO_2} \left( g \right) + \) ___ \( \ce{H_2O} \left( g \right) + 2,510 \; \text{kJ} \) As oxygen, O, is the only element that is present in this chemical formula, incorporating this coefficient does not alter the amounts of carbon, C, or hydrogen, H, that are present on the reactant side of the equation. The updated quantity of this element is reflected in the table that is shown below. Inserting this coefficient balances oxygen, O, as intended. Therefore, all of the components of this equation are now balanced. Element or Ion Reactants Products Balanced C 2 \( \cancel{\rm{1} } \) 2 NaN H 2 2 NaN O \( \bcancel{\rm{2} } \) 5 \( \cancel{\rm{3} } \) 5 NaN However, a fractional coefficient, 2.5 , is written in the equation that is shown above. As a result, all of the coefficients in this equation, including the unwritten "1"s that are understood to occupy the first and fourth blanks, must be multiplied by 2 , in order to cancel this half-fraction. The doubled coefficient values are reflected in the chemical equation that is shown below. 2 \(\ce{C_2H_2} \left( g \right) + \) 5 \( \ce{O_2} \left( g \right) \rightarrow\) 4 \( \ce{CO_2} \left( g \right) + \) 2 \( \ce{H_2O} \left( g \right) + 2,510 \; \text{kJ} \) By multiplying all of the coefficients in this equation by 2 , the quantities in which carbon, C, hydrogen, H, and oxygen, O, are present in the equation have changed, as shown in the table below, but their relative ratios have not. Therefore, all of the components of this equation are still balanced. Element or Ion Reactants Products Balanced C \( \cancel{\rm{2} } \) 4 \( \cancel{\rm{1} } \) \( \cancel{\rm{2} } \) 4 NaN H \( \cancel{\rm{2} } \) 4 \( \cancel{\rm{2} } \) 4 NaN O \( \bcancel{\rm{2} } \) \( \cancel{\rm{5} } \) 10 \( \cancel{\rm{3} } \) \( \cancel{\rm{5} } \) 10 NaN Finally, these coefficients cannot be divided, as they do not all share a common divisor that would result in the calculation of four whole number coefficients. Therefore, the final equation that is presented above is chemically-correct, as written. Answer b This reaction is classified as a combustion because an energetic quantity, " 2,510 kJ ," is included as a product. Furthermore, a combustion reaction requires that molecular oxygen, O 2 , participate in the reaction as a reactant and that carbon dioxide, CO 2 , and water, H 2 O, be generated as products. All of these chemicals are present in the given equation. Answer c Because the energetic quantity, "2,510 kJ," is written on the right side of the reaction arrow in the given chemical equation, heat is generated as a product . Therefore, this reaction is exothermic . Answer d Indicator Information & Equality Patterns The word " molecules " indicates that an Avogadro's number equality should be developed and applied to solve this problem. Furthermore, since " molecules " is an indicator word, this word is inserted as the second unit on the right side of this type of equality, and the formula for the chemical that is referenced in the given problem, "water," H 2 O, is incorporated into both of the secondary unit positions in this equality, as shown below. Because the given chemical information is the name of a covalent molecule, the indicator word " molecules " appropriately corresponds to the chemical formula that is applied in this equality. 1 mol H 2 O = 6.02 × 10 23 H 2 O molecules Additionally, because the chemical that is referenced in the problem, water, H 2 O, and an energetic quantity are both present in the given reaction equation, a stoichiometric equality should also be developed and applied to solve this problem. The left side of this equality should be created using the equality pattern that was presented and discussed in Section 4.26. Since stoichiometry studies the molar relationships that exist between chemicals and quantities that are involved in chemical reactions, the primary unit for this portion of the equality is "mol," and the chemical formula of water, H 2 O, is incorporated as the secondary unit on the left side of the equality that is being developed. Finally, since the coefficient that is associated with with water, H 2 O, is a " 2 " in the corresponding chemical equation, a 2 is inserted into the numerical position on the left side of this stoichiometric equality. Because this problem references an energetic quantity, rather than a second chemical that participates in the reaction, the amount of heat that is generated in the reaction is written on the right side of this stoichiometric equality, as shown below. 2 mol H 2 O = 2,510 kJ Dimensional Analysis In order to completely eliminate the given unit, "molecules of water," the Avogadro's number equality must be applied first. Specifically, the quantity on the right side of the Avogadro's number equality becomes the denominator in the first conversion factor that is applied to solve the given problem. While reversing the order of the two units on the right side of an Avogadro's number equality is not absolutely necessary, doing so more clearly illustrates the desired unit cancelation for this particular problem. The remaining portion of the equality is written in the numerator in the resultant conversion factor. However, the unit that results upon the cancelation of "molecules of water" is "mol H 2 O," which is not the desired final unit. Therefore, a conversion factor based on the stoichiometric equality must be applied as a second conversion factor. In order to completely cancel the intermediate unit "mol H 2 O," the quantity on the left side of this equality becomes the denominator in the second conversion factor that is applied to solve the given problem. The remaining portion of the equality is written in the numerator in the resultant conversion factor, as shown below. \( {\text {3.16}} \times {\text{10}^{24}}\) \({\cancel{\rm{molecules} \; \rm{H_2O}}} \times\) \( \dfrac{1 \; \bcancel{\rm{mol} \; \rm{H_2O}}}{6.02 \times 10^{23} \; \cancel{\rm{molecules} \; \rm{H_2O}}}\) × \( \dfrac{2,510 \; \rm{kJ}}{2 \; \bcancel{\rm{mol} \; \rm{H_2O}}}\) = \( {\text {6,587.7076...}}\) \({\rm{kJ}}\) ≈ \( {\text {6,590}}\) \({\rm{kJ}}\) The solution is calculated by multiplying the given number by the value in each numerator, and then dividing by the quantity in each denominator. When using a calculator, each conversion factor should be entered in parentheses, or the "=" key should be used after each division, and any quantity that is expressed in scientific notation should be offset by an additional set of parentheses. Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown above.