Without doing any calculations, determine the sign of ΔS_sys for each chemical reaction. a. 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) c. Na(s) + 2 Cl₂(g) → NaCl(s) d. N₂(g) + 3 H₂(g) → 2 NH₃(g)

Question

Accepted Answer

Hey, everyone. And welcome back to another video without doing any calculations determine the sign of the entropy change of the system. For each chemical reaction, a two moles of potassium chlorate decomposed into two moles of potassium chloride and three moles of oxygen gas. We are given to answer choices A positive and B negative. So first of all, let's remember that the entropy change of a system represents a disorder. It can be positive or negative. If we have a positive sign, it basically means that, that this order increases, right? If it's negative, it basically means that the disorder of our system decreases. And let's remember that the entropy change of formation of solids is the lowest of all the physical states followed by the entropy affirmation of liquids. And then we have gasses. In other words, gasses are the least ordered, right? And knowing that whenever we want to evaluate whether a specific reaction increases in entropy or decreases, we always want to focus on gasses because they are the most influential in our equations. So in this case, if we carefully look at the reaction, we notice that on the left hand side, we have zero modes of gasses because we have two moles of a solid. On the right hand side, we have three moles of gasses because we have three moles of oxygen. Now, we can essentially state that the amount of gasses increase as we go from a lower quantity or based on nothing to a greater quantity, which is three moles. So if the amount of gasses increases, it essentially means that the entropy change of the system is positive since we are forming a more disordered state out of a more ordered state, right. So in this case, we can conclude that the correct answer to this problem, it would be a, the entropy change of the system is positive because we're forming a greater amount of gasses compared to the reactant side. On the reactant side, we had no gasses. And on the product side, we had three mo of gasses are going from a lower quantity of gasses to a greater quantity of gasses. Thank you for watching.

Without doing any calculations, determine the sign of ΔS_sys for each chemical reaction. a. 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) c. Na(s) + 2 Cl₂(g) → NaCl(s) d. N₂(g) + 3 H₂(g) → 2 NH₃(g)

Verified Solution

Key Concepts

Entropy (ΔS)

States of Matter

Chemical Reaction Stoichiometry

Without doing any calculations, determine the sign of ΔSsys for each chemical reaction. a. 2 KClO3(s) → 2 KCl(s) + 3 O2(g) c. Na(s) + 2 Cl2(g) → NaCl(s) d. N2(g) + 3 H2(g) → 2 NH3(g)

Verified Solution

Key Concepts

Entropy (ΔS)

States of Matter

Chemical Reaction Stoichiometry

Without doing any calculations, determine the sign of ΔS_sys for each chemical reaction. a. 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g) c. Na(s) + 2 Cl₂(g) → NaCl(s) d. N₂(g) + 3 H₂(g) → 2 NH₃(g)