IB Chemistry home > Syllabus 2016 > Energetics > Two and three step process calculations

Syllabus ref: 5.2

Hess' law tells us that when a chemical reaction is broken down into two, or more stages, the overall energy change must be the sum of the individual steps. This fact can be used in conjunction with mathematical manipulation of equations and applied to find enthalpy change values for systems that cannot be measured experimentally.

Constructing equations

The key to solving energy problems is to construct the equation for the energy change required from other equations given in the question. The best practice is to see what is required for the final equation and add up the components one at a time from the equations given.

If we are required to find the enthalpy change for the following reaction:

2C(s) + 2H2(g) C2H4(g) ΔH = x kJ

and we are given the enthalpy of combustions of hydrogen, carbon and ethene as -285.8, -393.5 and -1141.1 kJ respectively.

Step 1 Write out all the equations for the information given:

1 H2(g) + ½O2(g) H2O(l)ΔH = -285.5 kJ
2 C(s) + O2(g) CO2(g)ΔH = -393.5 kJ
3 C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)    ΔH = -1141.1 kJ

Step 2 Inspect the equation to be constructed.

2C(s) + 2H2(g) C2H4(g) ΔH = x kJ

You need two moles of carbon on the left. In equation 2 there is only 1 mole of carbon.

Step 3 Multiply equation 2 by 2:

4 2C(s) + 2O2(g) 2CO2(g)ΔH = -787 kJ

We now need 2 moles of hydrogen.

Step 4 Multiply equation 1 by 2:

5 2H2(g) + O2(g) 2H2O(l)ΔH = -571 kJ

In our final equation the carbon and the hydrogen is added together.

Step 5 Add together these two new equations 4 and 5 to give equation 6 below:

6 2C(s) + 2H2(g) + 3O2(g) 2CO2(g) + 2H2O(l)      ΔH = -1358 kJ

Step 6 subtract equation 3 from equation 6 to remove 2CO2(g) + 2H2O(l) from the right hand side, and also place C2H4(g) on the left hand side as a negative item.

6 2C(s) + 2H2(g) + 3O2(g) 2CO2(g) + 2H2O(l)      ΔH = -1358 kJ
3 C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)    ΔH = -1141.1 kJ

7 2C(s) + 2H2(g) - C2H4(g) zero ΔH = -216.9 kJ

Step 7 Equation 7 can now be rearranged by taking the ethene to the right hand side:

8 2C(s) + 2H2(g) C2H4(g)ΔH = -216.9 kJ

Remember that this last rearrangement step does not affect the energy change.

Although the calculation may seem elaborate and long-winded, the procedure is always the same. Construct the desired equation in parts by multiplying (or dividing as appropriate) and then subtraction (or addition as appropriate).


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