Question

Explain with reason, sign convention of $\delta S$ in the following reaction.
$ { N }_{ 2 }(g)\quad +\quad { 3H }_{ 2 }(g)\quad \longrightarrow \quad 2{ NH }_{ 3 }(g)$

Answer 1

Hint: $S$ is the entropy. It is the measure of randomness and disorder in the system. $\delta S$ is referred to as the change in the entropy.

Complete step by step answer:
S is the measure of disorder in the thermodynamic system. As the degree of disorder increases, the positivity of S also increases. This means that greater the disorder in the thermodynamic system, the more positive will be the change in the entropy.

In the reaction given, we can see that the number of moles of the reactant is greater than the number of moles of the product.

$ { N }_{ 2 }(g)\quad +\quad { 3H }_{ 2 }(g)\quad \longrightarrow \quad 2{ NH }_{ 3 }(g)$
(total number of moles of gaseous reactants) > (total number of moles of gaseous product)
And we know that more the no. of moles, more is the disorder.
Therefore, (Disordered state of gaseous reactants) > (Disordered state of gaseous products)
In the reaction, we can see that 4 moles of gaseous reactants reacts to form 2 moles of gaseous products.

Therefore, change in the no. of moles, $\delta n >0$. And since $\delta n >0$, the disorder in the thermodynamic system decreases and with the decrease in the disorder, the entropy also decreases. Or we can say that $\delta S <0$. Or, the sign of $\delta S$ is negative.

Note: Don't confuse between $\delta S$ and $\delta H$. $\delta H$ is the change in enthalpy of the system, which is the total energy of the system. Whereas, $\delta S$ is the change in the entropy of the system, which is the randomness of the system.