Hess' Law

Key Questions

  • How do you compute Hess's law calculations?

    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.

    1. You can reverse the equation. This will change the sign of #ΔH#.
    2. You can multiply the equation by a constant. You must then multiply the value of #ΔH# by the same constant.
    3. You can use any combination of the first two rules.

    EXAMPLE:

    What is the value for the heat of combustion, #ΔH_c#, of the following reaction?

    #color(red)("CS"_2("l") + 3"O"_2("g") → "CO"_2("g") + 2"SO"_2("g"))#

    Given:

    #1. color(blue)("C"("s") + "O"_2("g") → "CO"_2(g); ΔH_f = "-393.5 kJ")#
    #2. color(blue)("S"("s") + "O"_2("g") → "SO"_2("g"); color(white)(l)ΔH_f = "-296.8 kJ")#
    #3. color(blue)("C"("s") + 2"S"("s") → "CS"_2("l"); color(white)(n)ΔH_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") → "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") → "CO"_2(g); ΔH_f = "-393.5 kJ")#
    #2. color(blue)("S"("s") + "O"_2("g") → "SO"_2("g"); color(white)(l)ΔH_f = "-296.8 kJ")#
    #3. color(blue)("C"("s") + 2"S"("s") → "CS"_2("l"); color(white)(n)ΔH_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 #ΔH# to get the #"CS"_2# on the left in Equation 4.

    #4. color(purple)("CS"_2("l") → "C"("s") + "2S"("s"); "-"ΔH_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") → "C"("s") + "2S"("s"); "-"ΔH_f = "-87.9 kJ")#
    #1. color(blue)("C"("s") + "O"_2("g") → "CO"_2(g); ΔH_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 #ΔH# to get Equation 5.

    We then get

    #4. color(purple)("CS"_2("l") → "C"("s") + "2S"("s"); "-"ΔH_f = "-87.9 kJ")#
    #1. color(blue)("C"("s") + "O"_2("g") → "CO"_2(g); ΔH_f = "-393.5 kJ")#
    #5. color(green)("2S"("s") + "2O"_2("g") → "2SO"_2("g"); ΔH_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") → cancel("C(s)") + cancel("2S(s)") color(white)(XXXXXlX)"-"ΔH_f = color(white)(n)"-87.9 kJ"#
    #cancel("C(s)") + "O"_2"(g)" → "CO"_2"(g)" color(white)(XXXXXXl)ΔH_f = "-393.5 kJ"#
    #cancel("2S(s)") + "2O"_2("g)" → "2SO"_2"(g)" color(white)(XXXXX)ΔH_f = "-593.6 kJ"#
    #stackrel("———————————————————————————————————")("CS"_2"(l)" + "3O"_2"(g)" → "CO"_2"(g)" + "2SO"_2"(g)"; ΔH_c = "-1075.0 kJ")#

    Answer:

    The heat of combustion for the reaction is -1075.0 kJ.

    Ernest Z. · 6 · Jan 11 2014
  • What does Hess's law say about the enthalpy of a reaction?

    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 (ΔH) 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, #ΔH_"c"#, of CS₂, given the following equations.

    1. C(s) + O₂(g) → CO₂(g); #ΔH_"c"# = -393.5 kJ
    2. S(s) + O₂(g) → SO₂(g); #ΔH_"c"# = -296.8 kJ
    3. C(s) + 2S(s) → CS₂(l); #ΔH_"f"# = 87.9 kJ

    Solution

    Write down the target equation, the one you are trying to get.

    CS₂(l) + 2O₂(g) → CO₂(g) + 2SO₂(g)

    Start with equation 3. It contains the first compound in the target (CS₂).

    We have to reverse equation 3 and its ΔH to put the CS₂ on the left. We get equation A below.

    A. CS₂(l) → C(s) + 2S(s); -#ΔH_"f"# = -87.9 kJ

    Now 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) + O₂(g) → CO₂(g); #ΔH_"c"# = -393.5 kJ

    We use Equation 2 to eliminate the S(s), but we have to double it to get 2S(s). We also double its #ΔH#. We then get equation C below.

    C. 2S(s) + 2O₂(g) → 2SO₂(g); #ΔH_"c"# = -593.6 kJ

    Finally, 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. CS₂(l) → C(s) + 2S(s); -#ΔH_"f"# = -87.9 kJ
    B. C(s) + O₂(g) → CO₂(g); #ΔH_"f"# = -393.5 kJ
    C. 2S(s) + 2O₂(g) → 2SO₂(g); #ΔH_"f"# = -593.6 kJ

    CS₂(l) + 3O₂(g) → CO₂(g) + 2SO₂(g); #ΔH_"c"# = -1075.0 kJ

    A. M. · 4 · Jun 22 2014
  • What is the Hess' law?

    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).

    Shivangi Mangotra · 5 · Oct 2 2015

Questions

Thermochemistry