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Now that we understand that chemical reactions occur with a simultaneous change in energy, we can apply the concept more broadly. To start, remember that some chemical reactions are rather difficult to perform. For example, consider the combustion of carbon to make carbon monoxide:
2C(s) + O_{2}(g) → 2CO(g) ΔH = ?In reality, this is extremely difficult to do; given the opportunity, carbon will react to make another compound, carbon dioxide:
2C(s) + O_{2}(g) → 2CO_{2}(g) ΔH = −393.5 kJIs there a way around this? Yes. It comes from the understanding that chemical equations can be treated like algebraic equations, with the arrow acting like the equals sign. Like algebraic equations, chemical equations can be combined, and if the same substance appears on both sides of the arrow, it can be canceled out (much like a spectator ion in ionic equations). For example, consider these two reactions:
2C(s) + 2O_{2}(g) → 2CO_{2}(g) 2CO_{2}(g) → 2CO(g) + O_{2}(g)If we added these two equations by combining all the reactants together and all the products together, we would get
2C(s) + 2O_{2}(g) + 2CO_{2}(g) → 2CO_{2}(g) + 2CO(g) + O_{2}(g)We note that 2CO_{2}(g) appears on both sides of the arrow, so they cancel:
$$\text{2C(s)}+{\text{2O}}_{\text{2}}\text{(g)}+\overline{){\text{2CO}}_{\text{2}}\text{(g)}}\to \overline{){\text{2CO}}_{\text{2}}\text{(g)}}+\text{2CO(g)}+{\text{O}}_{\text{2}}\text{(g)}$$We also note that there are 2 mol of O_{2} on the reactant side, and 1 mol of O_{2} on the product side. We can cancel 1 mol of O_{2} from both sides:
$$\text{2C(s)}+\overline{)\text{2}}{\text{O}}_{\text{2}}\text{(g)}\to \text{2CO(g)}+\overline{){\text{O}}_{\text{2}}\text{(g)}}$$What do we have left?
2C(s) + O_{2}(g) → 2CO(g)This is the reaction we are looking for! So by algebraically combining chemical equations, we can generate new chemical equations that may not be feasible to perform.
What about the enthalpy changes? Hess’s lawWhen chemical equations are combined algebraically, their enthalpies can be combined in exactly the same way. states that when chemical equations are combined algebraically, their enthalpies can be combined in exactly the same way. Two corollaries immediately present themselves:
What are the equations being combined? The first chemical equation is the combustion of C, which produces CO_{2}:
2C(s) + 2O_{2}(g) → 2CO_{2}(g)This reaction is two times the reaction to make CO_{2} from C(s) and O_{2}(g), whose enthalpy change is known:
C(s) + O_{2}(g) → CO_{2}(g) ΔH = −393.5 kJAccording to the first corollary, the first reaction has an energy change of two times −393.5 kJ, or −787.0 kJ:
2C(s) + 2O_{2}(g) → 2CO_{2}(g) ΔH = −787.0 kJThe second reaction in the combination is related to the combustion of CO(g):
2CO(g) + O_{2}(g) → 2CO_{2}(g) ΔH = −566.0 kJThe second reaction in our combination is the reverse of the combustion of CO. When we reverse the reaction, we change the sign on the ΔH:
2CO_{2}(g) → 2CO(g) + O_{2}(g) ΔH = +566.0 kJNow that we have identified the enthalpy changes of the two component chemical equations, we can combine the ΔH values and add them:
Hess’s law is very powerful. It allows us to combine equations to generate new chemical reactions whose enthalpy changes can be calculated, rather than directly measured.
Determine the enthalpy change of
C_{2}H_{4} + 3O_{2} → 2CO_{2} + 2H_{2}O ΔH = ?from these reactions:
C_{2}H_{2} + H_{2} → C_{2}H_{4} ΔH = −174.5 kJ 2C_{2}H_{2} + 5O_{2} → 4CO_{2} + 2H_{2}O ΔH = −1,692.2 kJ 2CO_{2} + H_{2} → 2O_{2} + C_{2}H_{2} ΔH = −167.5 kJSolution
We will start by writing chemical reactions that put the correct number of moles of the correct substance on the proper side. For example, our desired reaction has C_{2}H_{4} as a reactant, and only one reaction from our data has C_{2}H_{4}. However, it has C_{2}H_{4} as a product. To make it a reactant, we need to reverse the reaction, changing the sign on the ΔH:
C_{2}H_{4} → C_{2}H_{2} + H_{2} ΔH = +174.5 kJWe need CO_{2} and H_{2}O as products. The second reaction has them on the proper side, so let us include one of these reactions (with the hope that the coefficients will work out when all our reactions are added):
2C_{2}H_{2} + 5O_{2} → 4CO_{2} + 2H_{2}O ΔH = −1,692.2 kJWe note that we now have 4 mol of CO_{2} as products; we need to get rid of 2 mol of CO_{2}. The last reaction has 2CO_{2} as a reactant. Let us use it as written:
2CO_{2} + H_{2} → 2O_{2} + C_{2}H_{2} ΔH = −167.5 kJWe combine these three reactions, modified as stated:
What cancels? 2C_{2}H_{2}, H_{2}, 2O_{2}, and 2CO_{2}. What is left is
C_{2}H_{4} + 3O_{2} → 2CO_{2} + 2H_{2}Owhich is the reaction we are looking for. The ΔH of this reaction is the sum of the three ΔH values:
ΔH = +174.5 − 1,692.2 − 167.5 = −1,685.2 kJTest Yourself
Given the thermochemical equations
Pb + Cl_{2} → PbCl_{2} ΔH = −223 kJ PbCl_{2} + Cl_{2} → PbCl_{4} ΔH = −87 kJdetermine ΔH for
2PbCl_{2} → Pb + PbCl_{4}Answer
+136 kJ
Define Hess’s law.
What does Hess’s law require us to do to the ΔH of a thermochemical equation if we reverse the equation?
If the ΔH for
C_{2}H_{4} + H_{2} → C_{2}H_{6}is −65.6 kJ, what is the ΔH for this reaction?
C_{2}H_{6} → C_{2}H_{4} + H_{2}If the ΔH for
2Na + Cl_{2} → 2NaClis −772 kJ, what is the ΔH for this reaction:
2NaCl → 2Na + Cl_{2}If the ΔH for
C_{2}H_{4} + H_{2} → C_{2}H_{6}is −65.6 kJ, what is the ΔH for this reaction?
2C_{2}H_{4} + 2H_{2} → 2C_{2}H_{6}If the ΔH for
2C_{2}H_{6} + 7O_{2} → 4CO_{2} + 6H_{2}Ois −2,650 kJ, what is the ΔH for this reaction?
6C_{2}H_{6} + 21O_{2} → 12CO_{2} + 18H_{2}OThe ΔH for
C_{2}H_{4} + H_{2}O → C_{2}H_{5}OHis −44 kJ. What is the ΔH for this reaction?
2C_{2}H_{5}OH → 2C_{2}H_{4} + 2H_{2}OThe ΔH for
N_{2} + O_{2} → 2NOis 181 kJ. What is the ΔH for this reaction?
NO → 1/2N_{2} + 1/2O_{2}Determine the ΔH for the reaction
Cu + Cl_{2} → CuCl_{2}given these data:
2Cu + Cl_{2} → 2CuCl ΔH = −274 kJ 2CuCl + Cl_{2} → 2CuCl_{2} ΔH = −166 kJDetermine ΔH for the reaction
2CH_{4} → 2H_{2} + C_{2}H_{4}given these data:
CH_{4} + 2O_{2} → CO_{2} + 2H_{2}O ΔH = −891 kJ C_{2}H_{4} + 3O_{2} → 2CO_{2} + 2H_{2}O ΔH = −1,411 kJ 2H_{2} + O_{2} → 2H_{2}O ΔH = −571 kJDetermine ΔH for the reaction
Fe_{2}(SO_{4})_{3} → Fe_{2}O_{3} + 3SO_{3}given these data:
4Fe + 3O_{2} → 2Fe_{2}O_{3} ΔH = −1,650 kJ 2S + 3O_{2} → 2SO_{3} ΔH = −792 kJ 2Fe + 3S + 6O_{2} → Fe_{2}(SO_{4})_{3} ΔH = −2,583 kJDetermine ΔH for the reaction
CaCO_{3} → CaO + CO_{2}given these data:
2Ca + 2C + 3O_{2} → 2CaCO_{3} ΔH = −2,414 kJ C + O_{2} → CO_{2} ΔH = −393.5 kJ 2Ca + O_{2} → 2CaO ΔH = −1,270 kJIf chemical equations are combined, their energy changes are also combined.
ΔH = 65.6 kJ
ΔH = −131.2 kJ
ΔH = 88 kJ
ΔH = −220 kJ
ΔH = 570 kJ