Ethylene (CH2=CH2) represents one of the most valuable industrial chemicals. It is commonly used to synthesis styrene (C6H5C=CH2), the building block of polystyrene, a versatile plastic used to make a wide variety of consumer products. The overall chemical synthesis of styrene is given below: C6H6(l) + CH2=CH2(g) → C6H5C=CH2(l) + H2 (g) [table]
Compound | ∆Hf0 (298.15K), kJ mol-1 | S0 (298.15K), J K-1 mol-1 C6H6(l) | 48.7 | 173.26 CH2=CH2(g) | 52.47 | 219.32 C6H5C=CH2(l) | 108.0 | 240.5 H2 (g) | 0.0 | 114.7
[/table] What should happen to the entropy of the system during the reaction?

Question

BlackTom AI · Accepted Answer

The question asks: What should happen to the entropy of the system during the reaction?
First, I will establish the relevant data from the table: the reaction is
C6H6(l) + CH2=CH2(g) → C6H5C=CH2(l) + H2(g).
We have the standard molar entropies at 298 K:
- C6H6(l): S° = 173.26 J K−1 mol−1
- CH2=CH2(g): S° = 219.32 J K−1 mol−1
- C6H5C=CH2(l) (styrene liquid): S° = 240.5 J K−1 mol−1
- H2(g): S° = 114.7 J K−1 mol−1
To find the entropy change for the reaction, sum the product entropy values and subtract the sum of the reactant entropy values:
ΔSrxn = [S°(styrene l) + S°(H2 g)] − [S°(benzene l) + S°(ethylene g)]
= (240.5 + 114.7) − (173.26 + 219.32)
= 355.2 − 392.58
≈ −37.38 J K−1 mol−1.
So, the entropy of the system decreases slightly (negative ΔS).
Now, let’s evaluate the given answer option, even though the options list is incomplete here.
Option b. Decrease would be consistent with the calculated ΔSrxn ≈ −37 J K−1 mol−1, indicating a net loss of disorder in the system during this transformation.
Consider why this might be the case: although one mole of gas (H2) is produced, the reactants include benzene liquid and ethene gas, and the product is a single liquid (styrene) plus a gas. The net change in disorder is governed by the detailed S° values, which show that the product side is, in total, more ordered than the reactant side for this reaction.
Why the other (unshown) options would be incorrect in typical scenarios: if an option claimed an increase in entropy, it would contradict the calculated ΔSrxn using the provided standard entropies; if an option claimed no change, it would require ΔS ≈ 0, which is not supported by the numerical result; if an option asserted that entropy change is dominated by gas moles alone, that would ignore the actual S° values of the liquids and gases involved and could mislead about how entropy is distributed among translational, rotational, and vibrational modes.
In summary, based on the tabulated S° values for reactants and products, the reaction’s entropy change is negative, corresponding to a decrease in the system’s entropy.

CHM1051 MUM S1 2025 CHM1051 Practice Exam 1

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