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mrs_skeptik [129]
3 years ago
12

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)
Chemistry
1 answer:
prisoha [69]3 years ago
3 0

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

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