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Question
Answer the following question.
Although ΔS for the formation of two moles of water from H2 and O2 is –327J K–1, it is spontaneous. Explain.
(Given ΔH for the reaction is –572 kJ).
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Solution
1. For the process to be spontaneous, `triangle "S"_"total" = triangle "S"_"sys" + triangle "S"_"surr" > 0`.
2. For the reaction, 2H2(g) + O2(g) → 2H2O(l), when 2 moles of H2 and 1 mole of O2 gas combine to form 2 moles of liquid water, 572 kJ of heat is released which is received by surroundings at constant pressure (and 298 K).
3. The entropy change of the surroundings is,
`triangle "S"_"surr" = ("Q"_"rev")/"T" = (572 xx 10^3 "J")/(298 "K")` = + 1919 J K-1
4. The total enthalpy change is,
`triangle "S"_"total" = triangle "S"_"sys" + triangle "S"_"surr"`
= - 327 J K-1 + 1919 J K-1
= + 1592 J K-1
5. Since `triangle "S"_"total"` > 0, the reaction is spontaneous at 25 °C.
6. It follows that to decide spontaneity of reactions, we need to consider the entropy of system and its surroundings.
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