Question

Predict the sign of ΔS° for 2NO2(g) LaTeX: \longrightarrow⟶ N2O4(g) CaCO3(s) + 2HCl(aq) LaTeX: \longrightarrow⟶ CaCl2(aq) + H2O(l) +CO2(g) Ag+(aq) + Cl-(aq) LaTeX: \longrightarrow⟶ AgCl(s)

Answer 1

Answer: a. $2NO_2(g)\rightarrow N_2O_4(g)$: $\Delta S$ is negative

b. $CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)$ : $\Delta S$ is negative

c. $Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)$: $\Delta S$ is negative

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa

a)  $2NO_2(g)\rightarrow N_2O_4(g)$

In this reaction 2 moles of gaseous reactants are converting to 1 mole of gaseous products. The randomness will decrease and hence entropy will also decrease. Thus $\Delta S$ is negative.

b) $CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)$

In this reaction solid reactants are converting to aqueous products. The randomness will increase and hence entropy will also increase. Thus $\Delta S$ is positive.

c) $Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)$

In this reaction aqueous reactants are converting to solid products. The randomness will decrease and hence entropy will also decrease. Thus $\Delta S$ is negative