To understand how to use Hess\'s law to find the enthalpy of an overall reaction

Question

To understand how to use Hess's law to find the enthalpy of an overall reaction. The change in enthalpy, delta H, is the heat absorbed or produced during any reaction at constant pressure. Hess's law states that delta H for an overall reaction is the sum of the delta H values for the individual reactions. For example, if we wanted to know the enthalpy change for the reaction 3Mn + 3O_2 rightarrow 3MnO_2 we could calculate it using the enthalpy values for the following individual steps: Step 1: 4Al + 3O_2 rightarrow 2Al_2O_3 Step 2: 3Mn + 2Al_2O_3 rightarrow 3MnO_2 + 4Al Overall: 3Mn + 3O_2 rightarrow 3MnO_2 If the enthalpy change is -3352 kJ/mol for step 1 and 1792 kJ/mol for step 2, then the enthalpy change for the overall reaction is calculated as follows: delta H = -3352 + 1792 = -1560 kJ/mol It is also important to note that the change in enthalpy is a state function, meaning it is independent of path. In other words, the sum of the delta H values for any set of reactions that produce the desired product from the starting materials gives the same overall delta H. Now consider the following set of reactions: C_2H_5OH + 3O_2 rightarrow 2CO_2 + 3H_2O, delta H = -1370 kJ/mol C_2H_4 + 3O_2 rightarrow 2CO_2 + 2H_2O, delta H = -1410 kJ/mol The equations given in the problem introduction can be added together to give the following reaction: overall: C_2H_4 + H_2O rightarrow C_2H_5OH However, one of them must be reversed. Which one? reaction 1: C_2H_5OH + 3O_2 rightarrow 2CO_2 + 3H_2O reaction 2: C_2H_4 + 3O_2 rightarrow 2CO_2 + 2H_2O What is the enthalpy for reaction 1 reversed? reaction 1 reversed: 2CO_2 + 3H_2O rightarrow C_2H_5OH + 3O_2 Express your answer numerically in kilojoules per mole.

Explanation / Answer

Part A : The reaction to be reversed

reaction 1

Part B : enthalpy of reaction 1 reversed

dH = +1370 kJ/mol

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