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 tor 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 +2Al_2O_3 Step 2: 3Mn + 2Al_2O_2 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 lot 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 of the delta H values for any set of reactions that produce the desired product from the staring materials gives the same overall delta H. However, one of them must be reversed. Which one? reactions 1: CO + 1/2 O_2 rightarrow CO_2 reactions 2: C + O_2 rightarrow CO_2 What is the enthalpy for reaction f reversed? reaction 1 reversed CO_2 rightarrow CO + 1/2 O_2 Express your answer numerically in kilojoules per mole.

Explanation / Answer

first provide H of reaction 1 and eaction 2

there are no given so we take normal value of   H

2. C(s) + O2 ---> CO2 H = -393.5 KJ/mole
1. CO + 1/2 O2 ---> CO2 H = -283 KJ/mole

To get the equation C(s)+ 1/2 O2 ---> CO from these 2, we must flip the 1 equation as such:

2. C(s) + O2 ---> CO2 H = -393.5 KJ/mole
1. CO2 ---> CO + 1/2 O2 H = +283 KJ/mole {reaction will reverse then H will reverse sign}

Crossing out the CO2 and 1/2 O2 from products and reactants give us:
C(s)+ 1/2 O2 ---> CO
And so we just need to add the half enthalpies to get (-393 KJ + 283 KJ ) = -110 kJ/mole

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