P 4. Review the Knowledge You Need to Score High Free-Response Question You have
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P 4. Review the Knowledge You Need to Score High Free-Response Question You have 10 minutes to answer the following question. You may use a calculator and the tables in the back of the book Question1 Xe(g) + 3F2 (g) XeFg Under standard conditions, the enthalpy change for the reaction going from left to right (a) Is the value of AS, for the above reaction, positive or negative? Justify your conclusion. (b) The above reaction is spontaneous under standard conditions. Predict what will happen (forward reaction) is AH294 k). to ? G for this reaction as the temperature is increased. Justify your prediction. (c) Will the value of K remain the same, increase, or decrease as the temperature increases? (d) Show how the temperature at which the reaction changes from spontaneous to non- Justify your prediction. spontaneous can be predicted. What additional information is necessary?Explanation / Answer
The given reaction is
Xe (g) + 3 F2 (g) <=====> XeF6 (g)
a) The standard entropy change of a reaction, ?S0 depends on the number of gaseous molecules present in the system. The entropy, S° of a system is a measure of the randomness of the system. The higher the number of gaseous molecules, the higher will be the random motions and hence, the higher the entropy. Since the reaction proceeds with a decrease in the number of gaseous molecules (4 on the reactant side, 1 on the product side), hence the entropy of the product(s) is/are lower than the entropy of the reactants and consequently, ?S0 < 0, i.e, ?S0 is negative.
b) We know that
?G = ?H0 – T*?S0
Under standard conditions (room temperature, T = 298 K), the reaction is spontaneous, i.e, ?G0 < 0. Since ?H0 and ?S0 are both negative, ?G0 is negative only when ?H0 >> T*?S0. As T is increased, T*?S0 increases (becomes more negative) and approaches a condition when ?H0 = T*?S0. At that condition, ?G = 0. When T is increased further, T*?S0 > ?H0 and ?G > 0, i.e, the reaction becomes non-spontaneous.
c) K is a thermodynamic equilibrium constant and depends on the temperature. As temperature increases, K increases.
d) We can find out the temperature when the reaction becomes non-spontaneous. We need to know the numerical values of both ?H0 and ?S0 and use the condition ?G = 0. Therefore,
?H0 = T’*?S0
=====> T’ = ?H0/?S0
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