Question

Part A

Predict the sign of the entropy change, ΔS∘, for each of the reaction displayed.

Drag the appropriate items to their respective bins.

Ag+(aq)+Cl−(aq)→AgCl(s)

2KClO3(s)→2KCl(s)+3O2(g)

2N2O(g)→2N2(g)+O2(g)

2Mg(s)+O2(g)→2MgO(s)

C7H16(g)+11O2(g)→7CO2(g)+8H2O(g)

H2O(l)→H2O(g)

Part B

Calculate the standard entropy change for the reaction

2Mg(s)+O2(g)→2MgO(s)

using the data from the following table:

Substance	ΔH∘f (kJ/mol)	ΔG∘f (kJ/mol)	S∘ [J/(K⋅mol)]
Mg(s)	0.00	0.00	32.70
O2(g)	0.00	0.00	205.0
MgO(s)	-602.0	-569.6	27.00

Express your answer to four significant figures and include the appropriate units.

ΔS∘ =

Answer 1

Entropy is defined as:

Entropy = Entropy of products - Entropy of reactants

negative entropy: (Entropy of reactants greater than products)

Ag⁺_(aq)+Cl⁻_(aq)→AgCl_(s)

2Mg(s)+O2(g)→2MgO(s)

Positive entropy: (Entropy of products is greater than reactants)

2KClO_3(s)→2KCl_(s)+3O_2(g)

2N₂O(g)→2N₂(g)+O₂(g)

C₇H₁₆(g)+11O₂(g)→7CO₂(g)+8H₂O(g)

H₂O(l)→H₂O(g)

Part B

Entropy of reaction:

2Mg(s)+O2(g)→2MgO(s)

Entropy of products

MgO : 27, 2 MgO = 2 * 27 = 54 J / K mol

Entropy of reactants

2 Mg = 32.7 J / K mol

O₂ = 205 J / K mol

Entropy of reaction : 54 - (32.7+205) = -183.7 J / K mol

*hope it helps =)