# Question 62075

Mar 9, 2017

Hess's Law is saying that if you convert reactants A into products B, the overall enthalpy change will be exactly the same whether you do it in one step or two steps or however many steps.

$\Delta H = - 254 k \frac{J}{m o l}$

#### Explanation:

Hess’s Law states “The enthalpy change accompanying a chemical change is independent of the route by which the chemical change occurs.”

Hess's Law is saying that if you convert reactants A into products B, the overall enthalpy change will be exactly the same whether you do it in one step or two steps or however many steps. http://chemguide.co.uk/physical/energetics/sums.html

This is extremely useful in calculating the types of reaction information that your question illustrates. By setting up “equations” of the known reactions in a way to generate the desired one, a thermodynamic value for the desired reaction can be determined.

The energy of formation of propanone is the sum the energy of decomposition of its constituent parts from that of the overall molecule. Write out each combustion reaction, then combine them into the equivalent one for propanone and solve the algebra.

2H_2 + O_2 → 2H_2O ; –286 kJ/(mol)
C + O_2 → CO_2 ; –394 kJ/(mol)
C_3H_6O + 4O_2 → 3CO_2 + 3H_2O ; –1786 kJ/(mol)
Delta H – 1786 = 3*(-394) + 3*(-286)
$\Delta H = 1786 - 3 \cdot \left(- 394\right) + 3 \cdot \left(- 286\right)$
Delta H = 1786 - 1182 – 858 = -254 kJ/(mol)#