Hess' Law
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Key Questions

Answer:
According to the Hess's Law of constant heat summation, the total amount of heat evolved or absorbed in a reaction is same whether reaction takes place in one step or multiple steps.
Explanation:
All chemical reactions that take place around us might not be using heat energy always for there completion but there are some reactions which account to heat energy for there completion and use the same amount of heat energy if we complete the reaction process only in one step or in multiple number of steps.
FOR EXAMPLE, FORMATION OF CARBON DIOXIDE
The formation of carbon dioxide can take place in two steps which are given below: 
Firstly, we can directly react 1 mole of carbon with 1 molecule of oxygen we will give 1 mole of carbon dioxide. As, this reaction is an exothermic reaction there will be a liberation of 393.5 KJ/mol of heat energy.

OR we can break this whole reaction process into two parts:
1. To the first part we can combine 1 mole of carbon with half mole of oxygen molecule which will lead to the formation of 1 mol of carbon monoxide (CO) with the liberation of 110.5 KJ/mole of heat energy.
2. To the second part we can combine 1 mole of the formed carbon monoxide with half mole of an oxygen molecule which will lead to the formation of exactly 1 mole of carbon dioxide with the liberation of 283.0 KJ/mol of heat energy.
In total this two part reaction will also liberate  393.5 KJ/mol of heat energy which is exactly the same amount of heat energy that was liberated when we performed the reaction process directly in one step.
(The ve sign used above indicates the liberation of heat energy).
 FORMATION OF CARBON DIOXIDE

Answer:
You have to develop a strategy for the order in which you add the various equations.
Explanation:
Hess's law states that the total enthalpy change does not rely on the path taken from beginning to end.
So, you can calculate the enthalpy as the sum of several small steps.
There are a few rules that you must follow when manipulating an equation.
 You can reverse the equation. This will change the sign of
#Î”H# .  You can multiply the equation by a constant. You must then multiply the value of
#Î”H# by the same constant.  You can use any combination of the first two rules.
EXAMPLE:
What is the value for the heat of combustion,
#Î”H_c# , of the following reaction?#color(red)("CS"_2("l") + 3"O"_2("g") â†’ "CO"_2("g") + 2"SO"_2("g"))# Given:
#1. color(blue)("C"("s") + "O"_2("g") â†’ "CO"_2(g); Î”H_f = "393.5 kJ")#
#2. color(blue)("S"("s") + "O"_2("g") â†’ "SO"_2("g"); color(white)(l)Î”H_f = "296.8 kJ")#
#3. color(blue)("C"("s") + 2"S"("s") â†’ "CS"_2("l"); color(white)(n)Î”H_f = color(white)(X)"87.9 kJ")# Solution:
Write down the target equation (the one you are trying to get).
#color(red)("CS"_2("l") + 3"O"_2("g") â†’ "CO"_2("g") + 2"SO"_2("g"))# Write down the three equations you must use to get the target equation.
#1. color(blue)("C"("s") + "O"_2("g") â†’ "CO"_2(g); Î”H_f = "393.5 kJ")#
#2. color(blue)("S"("s") + "O"_2("g") â†’ "SO"_2("g"); color(white)(l)Î”H_f = "296.8 kJ")#
#3. color(blue)("C"("s") + 2"S"("s") â†’ "CS"_2("l"); color(white)(n)Î”H_f = color(white)(X)"87.9 kJ")# Now we need to organize the given equations so that they add up to give the target equation.
A good place to start is to find one of the equations that contains the first compound in the target equation (
#"CS"_2# ) .That would be equation 3, but we must reverse equation 3 and its
#Î”H# to get the#"CS"_2# on the left in Equation 4.#4. color(purple)("CS"_2("l") â†’ "C"("s") + "2S"("s"); ""Î”H_f = "87.9 kJ")# This equation contains
#"C"("s")# and#"S"("s")# , neither of which is in the target equation.We have to eliminate these one at a time. First, we find an equation that contains
#"C"("s")"# .That would be equation 1, since we have already used equation 3.
At this point, we have
#4. color(purple)("CS"_2("l") â†’ "C"("s") + "2S"("s"); ""Î”H_f = "87.9 kJ")#
#1. color(blue)("C"("s") + "O"_2("g") â†’ "CO"_2(g); Î”H_f = "393.5 kJ")# Now we work on the
#"S"("s")# .We will use equation 2, but we will have to double it and its
#Î”H# to get Equation 5.We then get
#4. color(purple)("CS"_2("l") â†’ "C"("s") + "2S"("s"); ""Î”H_f = "87.9 kJ")#
#1. color(blue)("C"("s") + "O"_2("g") â†’ "CO"_2(g); Î”H_f = "393.5 kJ")#
#5. color(green)("2S"("s") + "2O"_2("g") â†’ "2SO"_2("g"); Î”H_f = "593.6 kJ")# Finally, we add the three equations to get the target equation, cancelling things that appear on opposite sides of the reaction arrows.
#"CS"_2("l") â†’ cancel("C(s)") + cancel("2S(s)") color(white)(XXXXXlX)""Î”H_f = color(white)(n)"87.9 kJ"#
#cancel("C(s)") + "O"_2"(g)" â†’ "CO"_2"(g)" color(white)(XXXXXXl)Î”H_f = "393.5 kJ"#
#cancel("2S(s)") + "2O"_2("g)" â†’ "2SO"_2"(g)" color(white)(XXXXX)Î”H_f = "593.6 kJ"#
#stackrel("â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”â€”")("CS"_2"(l)" + "3O"_2"(g)" â†’ "CO"_2"(g)" + "2SO"_2"(g)"; Î”H_c = "1075.0 kJ")# Answer:
The heat of combustion for the reaction is 1075.0 kJ.
 You can reverse the equation. This will change the sign of

The law states that the total enthalpy change during a reaction is the same whether the reaction is made in one step or in several steps.
In other words, if a chemical change takes place by several different routes, the overall enthalpy change is the same, regardless of the route by which the chemical change occurs (provided the initial and final condition are the same).
Hess' law allows the enthalpy change (Î”H) for a reaction to be calculated even when it cannot be measured directly. This is accomplished by performing basic algebraic operations based on the chemical equation of reactions using previously determined values for the enthalpies of formation.
Addition of chemical equations leads to a net or overall equation. If enthalpy change is known for each equation, the result will be the enthalpy change for the net equation.
EXAMPLE
Determine the heat of combustion,
#Î”H_"c"# , of CSâ‚‚, given the following equations. C(s) + Oâ‚‚(g) â†’ COâ‚‚(g);
#Î”H_"c"# = 393.5 kJ  S(s) + Oâ‚‚(g) â†’ SOâ‚‚(g);
#Î”H_"c"# = 296.8 kJ  C(s) + 2S(s) â†’ CSâ‚‚(l);
#Î”H_"f"# = 87.9 kJ
Solution
Write down the target equation, the one you are trying to get.
CSâ‚‚(l) + 2Oâ‚‚(g) â†’ COâ‚‚(g) + 2SOâ‚‚(g)
Start with equation 3. It contains the first compound in the target (CSâ‚‚).
We have to reverse equation 3 and its Î”H to put the CSâ‚‚ on the left. We get equation A below.
A. CSâ‚‚(l) â†’ C(s) + 2S(s); 
#Î”H_"f"# = 87.9 kJNow we eliminate C(s) and S(s) one at a time. Equation 1 contains C(s), so we write it as Equation B below.
B. C(s) + Oâ‚‚(g) â†’ COâ‚‚(g);
#Î”H_"c"# = 393.5 kJWe use Equation 2 to eliminate the S(s), but we have to double it to get 2S(s). We also double its
#Î”H# . We then get equation C below.C. 2S(s) + 2Oâ‚‚(g) â†’ 2SOâ‚‚(g);
#Î”H_"c"# = 593.6 kJFinally, we add equations A, B, and C to get the target equation. We cancel things that appear on opposite sides of the reaction arrows.
A. CSâ‚‚(l) â†’ C(s) + 2S(s); 
#Î”H_"f"# = 87.9 kJ
B. C(s) + Oâ‚‚(g) â†’ COâ‚‚(g);#Î”H_"f"# = 393.5 kJ
C. 2S(s) + 2Oâ‚‚(g) â†’ 2SOâ‚‚(g);#Î”H_"f"# = 593.6 kJCSâ‚‚(l) + 3Oâ‚‚(g) â†’ COâ‚‚(g) + 2SOâ‚‚(g);
#Î”H_"c"# = 1075.0 kJ  C(s) + Oâ‚‚(g) â†’ COâ‚‚(g);
Questions
Videos on topic View all (7)
Thermochemistry

1Energy Change in Reactions

2Enthalpy

3Exothermic processes

4Specific Heat

5Calorimetry

6Thermochemistry of Phase Changes

7Thermochemistry with Equation Stoichiometry

8Hess' Law

9Spontaneous and NonSpontaneous Processes

10Entropy

11Gibbs Free Energy

12Endothermic processes

13BornHaber Cycle  Formation

14BornHaber Cycle  Solution