Without doing any calculations, determine the sign of S sys for each of the foll

Question

Without doing any calculations, determine the sign of Ssys for each of the following chemical reactions.

Part A

2KMnO4(s)K2MnO4(s)+MnO2(s)+O2(g)

Part B

CH2=CH2(g)+H2(g)CH3CH3(g)

Part C

Na(s)+1/2Cl2(g)NaCl(s)

Part D

2NO(g)+O2(g)2NO2(g)

Without doing any calculations, determine the sign of Ssys for each of the following chemical reactions.

Part A

2KMnO4(s)K2MnO4(s)+MnO2(s)+O2(g)

-Ssys>0 -Ssys<0

Part B

CH2=CH2(g)+H2(g)CH3CH3(g)

-Ssys>0 -Ssys<0

Part C

Na(s)+1/2Cl2(g)NaCl(s)

-Ssys<0 -Ssys>0

Part D

2NO(g)+O2(g)2NO2(g)

-Ssys<0 -Ssys>0

Explanation / Answer

The entropy of a system can be determined by looking at the states of the molecules like the entropy of gasses will be more than liquid and liquid will have more entropy than solid.

Part A

2KMnO4(s)K2MnO4(s)+MnO2(s)+O2(g)

Ssys > 0

Ssys = S(product) - S(reactant)

so for this part the only gas molecule is oxygen in that too is in product side so the entropy will higher for right side and

Ssys > 0

Part B

CH2=CH2(g)+H2(g)CH3CH3(g)

Ssys < 0

Part C

Na(s)+1/2Cl2(g)NaCl(s)

Ssys < 0

Part D

2NO(g)+O2(g)2NO2(g)

Ssys < 0

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