How do you compute Hess's law calculations?

1 Answer

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.