Question

How do you compute Hess’s law calculations?

Answer 1

Hint:Hess’s law states that the change in enthalpy in a chemical reaction is the same whether the process is carried out in one step or in multiple steps. It is the outcome of the first law of thermodynamics. It is used for calculation of standard reaction enthalpy.

Complete step-by-step answer:Hess law is used for calculating the enthalpies of neutralization for acid base reaction.
Now let us see how hess law is applied for the formation if sulphur trioxide-
formation of sulphur dioxide gas
$S+{{O}_{2}}\to S{{O}_{2}}$
$\Delta {{H}_{1}}=-70.96KCalmo{{l}^{-1}}$
now converting the sulphur dioxide gas into sulphur trioxide
$S{{O}_{2}}+\dfrac{1}{2}{{O}_{2}}\to S{{O}_{3}}$
$\Delta {{H}_{2}}=-23.49KCalmo{{l}^{-1}}$
The sulfur dioxide is removed from both the reaction and therefore when sulphur and oxygen react it results in the formation of sulphur trioxide.
Now using hess law we calculate standard reaction enthalpy-
The general formula is –
$\Delta {{H}_{R}}=\Delta {{H}_{1}}+\Delta {{H}_{2}}+\Delta {{H}_{3}}+\Delta {{H}_{4}}..........$
In the question,
$\Rightarrow \Delta {{H}_{R}}=\Delta {{H}_{1}}+\Delta {{H}_{2}}$
$\Rightarrow \Delta {{H}_{R}}=\left( -23.49 \right)+\left( -70.96 \right)$
$\Rightarrow \Delta {{H}_{R}}=-94.95KCalmo{{l}^{-1}}$
The net reaction for the formation of sulphur trioxide is-
$S+\dfrac{3}{2}{{O}_{2}}\to S{{O}_{3}}$
$\Delta {{H}_{R}}=-94.95KCalmo{{l}^{-1}}$
Additional information:The hess law cycle is the representation of formation of reactants and product from their respective elements in the standard state.
The reactants react and form a product in a single step or in multiple steps.

Note:There are many applications of hess law of heat summation-
When enthalpy changes in physical change- the allotropes of carbon that is graphite and diamond, the enthalpy change is calculated when graphite is converted into diamond.
When enthalpy changes in chemical reaction- if enthalpy of formation is negative then exothermic reaction takes place and if the enthalpy of formation is positive, endothermic reaction takes place. The bond energy is different for all the atoms and therefore it will show different enthalpy change in different products.