Final Answer-Explanation:
To calculate the standard entropy change (∆S°) for the given reaction, we'll use the following equation:
∆S° = ΣS°(products) - ΣS°(reactants)
First, we need to find the values of standard entropies for the reactants and products from Table 18.1.
From Table 18.1:
For C2H4(g), the standard molar entropy (S°) is 219 J/mol*K
For O2(g), the standard molar entropy (S°) is 205 J/mol*K
For CO2(g), the standard molar entropy (S°) is 213 J/mol*K
For H2O(l), the standard molar entropy (S°) is 188 J/mol*K
Now, we can calculate ∆S°:
∆S° = (2 * S°(CO2) + 2 * S°(H2O)) - (S°(C2H4) + 3 * S°(O2))
∆S° = (2 * 213 J/mol*K + 2 * 188 J/mol*K) - (219 J/mol*K + 3 * 205 J/mol*K)
∆S° = (426 J/mol*K + 376 J/mol*K) - (219 J/mol*K + 615 J/mol*K)
∆S° = 802 J/mol*K - 834 J/mol*K
∆S° = -32 J/mol*K
The calculated ∆S° for the reaction is -32 J/mol*K. Since ∆S° is negative, it indicates that the entropy of the system decreases during the reaction. This means that the products have lower entropy compared to the reactants.
In summary, the standard entropy change (∆S°) for the given reaction is -32 J/mol*K, and the entropy of the chemical system is expected to decrease as the reaction proceeds. This is in line with the negative value of ∆S°, indicating a decrease in entropy for the system.
