# Predict the sign of the entropy change of the system

Predict the sign of the entropy change of the system for each of the following reactions:
${N}_{2}\left(g\right)+3{H}_{2}\left(g\right)⇒2N{H}_{3}\left(g\right)$
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Neil Dismukes
Step 1
The reaction is:
${N}_{2}\left(g\right)+3{H}_{2}\left(g\right)⇒2N{H}_{3}\left(g\right)$
Based on the reaction, there are less gaseous component in the product side than in thereactant side. Thus, the final entropy must be less than the initial entropy. Hence, the change in entropy is negative.

Toni Scott
Step 1
$\mathrm{\Delta }S=\text{Numbr of moles of gaseous product}-\text{number of moles of gaseous reactant}.$
Based on above expression, we can say that sign of the change in entropy of system will be positive if the number of moles of gaseous product is greater than the number of moles of gaseous reactant and vice versa.
Given reaction:
${N}_{2}\left(g\right)+3{H}_{2}\left(g\right)⇒2N{H}_{3}\left(g\right)$

Therefore, sign of the given reaction is negative.

nick1337

Step 1
A decrease in entropy.
Entropy is associated with the degree of disorder, or number of degrees of freedom. In the reaction ${N}_{2}+3{H}_{2}-⇒2N{H}_{3}$ we start with four moles of gaseous reactants! and end with two moles of gaseous products, which is less disordered.
In general, when a reaction has fewer moles of gaseous products than the moles of gaseous reactants, this will be reflected in a decrease in entropy, and vice versa. It's a good job the enthalpy change for this reaction is exothermic, or it would be impossible to make ammonia from nitrogen and hydrogen.