Question

The standard enthalpy of decomposition of the yellow complex ${{H}_{3}}NS{{O}_{2}}$ into $N{{H}_{3}}$ and $S{{O}_{2}}$ is $+40kJmo{{l}^{-1}}$. Calculate the standard enthalpy of formation of ${{H}_{3}}NS{{O}_{2}},\Delta H_{f}^{0}(N{{H}_{3}}) = -46.17kJmo{{l}^{-1}},$$\Delta H_{f}^{0}(S{{O}_{2}}) = -296.83kJmo{{l}^{-1}}$

Answer 1

Hint: This question is an application of the third law of thermodynamics.
We could find the standard enthalpy of formation by the equation, $383kJmo{{l}^{-1}}$ \[\text{Enthalpy of reaction = enthalpy of products - enthalpy of reactants}\]

Complete Solution :
Here in the question it is given that a yellow complex, which is ${{H}_{3}}NS{{O}_{2}}$ and it is undergoing decomposition to form the products such as ammonia and sulphur dioxide.
And we are asked to find out the enthalpy of formation of the yellow complex, if the enthalpy of formation of $N{{H}_{3}}$ and $S{{O}_{2}}$.
- This term enthalpy is very familiar as it is a parameter widely discussed in thermodynamics.
From the third law of thermodynamics we know that energy can be neither created nor destroyed during a reaction, the energy is only inter-convertible and is conserved in a reaction.
So here, the value for standard enthalpy for the decomposition of the given complex will be equal to the value for the standard enthalpy of formation of the complex from $N{{H}_{3}}$ and $S{{O}_{2}}$.
And hence we can write the equation for this reaction of formation of ${{H}_{3}}NS{{O}_{2}}$ as,
$N{{H}_{3}}+S{{O}_{2}}\rightleftarrows {{H}_{3}}NS{{O}_{2}}$

- Now let’s write the data provided in the question.
The standard enthalpy value for the decomposition of ${{H}_{3}}NS{{O}_{2}}$ = $+40kJmo{{l}^{-1}}$
The value for enthalpy of formation of ammonia, $\Delta H_{f}^{0}(N{{H}_{3}}) = -46.17kJmo{{l}^{-1}}$
The value for the enthalpy of formation of sulphur dioxide, $\Delta H_{f}^{0}(S{{O}_{2}}) = -296.83kJmo{{l}^{-1}}$\[\text{Enthalpy of reaction = enthalpy of products - enthalpy of reactants}\]
Let’s substitute the values in the equation, here we take the heat of the reactant as the standard enthalpy of formation of ${{H}_{3}}NS{{O}_{2}}$.
Then we get,
\[\text{+40= (}-46.17)+\left( -296.83 \right)\text{-enthalpy of reactants}\]
\[\text{Enthalpy of reactants = -40+ (}-46.17)+\left( -296.83 \right) = 383kJmo{{l}^{-1}}\]
So the enthalpy for the formation of, ${{H}_{3}}NS{{O}_{2}} = 383kJmo{{l}^{-1}}$

Note: We should take care while doing the mathematical operations with the integers, there is a high chance of making mistakes in the calculation.
- While doing the calculation not only the magnitude but also the sign should also be taken into account.
- If in the question it is said that the reaction is an endothermic reaction then that enthalpy value possesses a positive sign and for an exothermic product, the value will have a negative sign.