Question

Calculate the standard enthalpy change for the reaction2A+B⇌2C+2D where the heats of formation are given in the following table:Substance ΔH∘f ( kJ/mol )A -275 B -389 C 185 D -519Express your answer in kilojoules.

Answer 1

`\Delta H\degree_(rxn)=271\text{ kJ}`

Procedure

To solve this problem you should start by writing down the reaction.

2A+B⇌2C+2D

Now, we will assume that the values represent the respective standard enthalpies of formation for the species that take part in the reaction. Standard enthalpies of formation are measured when one mole of each of these compounds is formed from its constituent elements in their pure state.

In your case, we will have:

A → -275

B →-389

C →185

D →-519

You can calculate the standard enthalpy change of a reaction using the Hess Law that expresses the enthalpy change of reaction by using each individual reaction that corresponds to the standard enthalpy change of formation for the products and for the reactants, as follows:

`\Delta H\degree_(rxn)=\Sigma(n*\Delta H_{f\mathring{pr}oducts})-\Sigma(m*\Delta H\degree_(freactants))`

Therefore for our case it will look as

`\Delta H\degree_(rxn)=\Sigma(2\text{ mol}*\frac{185\text{ kJ}}{mol}+2*\frac{-519\text{ kJ}}{mol})-\Sigma(2\text{ mol}*\frac{-275\text{ kJ}}{mol}+1\text{ }mol\frac{-389\text{ kJ}}{mol})`

This gives us as result

`\Delta H\degree_(rxn)=271\text{ kJ}`