Lesson: Chapter - 9

Enthalpies of Reactions

You can find the overall enthalpy of a reaction by subtracting the enthalpy at the beginning of the reaction from the enthalpy at the end of the reaction:

DH = H_final- H_initial
which is virtually the same as saying
DH = H_products -H_reactants

under the thermodynamic standard states 25ºC (298K), 1 atm, and 1 M. Try using this equation in a problem.

Example

Calculate the DH for the following:
3Al_(s) + 3NH₄ClO_4(s) ? Al₂O_3(s)+ AlCl_3(s) + 3NO_(g) + 6H₂O_(g)
given the following values:

Explanation

[1(-1676) + 1(-704) + 3(90) + 6(-242)] - [3(0) + 3(-295)] = -2677 kJ

This is an exothermic reaction since DH is negative. Don’t forget to multiply values by coefficients since each coefficient represents the number of moles of each substance!

Try another one:

Example

Find the DH_f of C₆H₁₂O_6(s) using the following information:

Explanation

[6(-393.5) + 6(-285.8)] - [1(x) + 6(0)] = -2800 kJ

Now solve the above for x and you have your answer! The value 2800 is negative because this reaction is exothermic. Oxygen is considered an element in its free state, so it is assigned a value of zero. (All diatomic molecules are assigned zeros for the same reason.) After solving for x, you get

for glucose.

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