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Calculate the entropy change in surround...

Calculate the entropy change in surroundings when 1.00 mol of `H_(2)O_((l))` is formed under standard conditions. `Delta_(f)H^(theta)=-286 kJ mol^(-1)`

Text Solution

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The correct Answer is:
Entropy change in surroundings `=959.73 JK^(-1).`
Given: `" "H_(2)(g) + (1)/(2)O_(2)(g) rarr H_(2)O(l), Delta_(r)H^(@) =- 286 KJ mol^(-1)`
Asked : The entropy change for the reaction.=?
Formula:`" "DeltaS_("surr") = (qsurr)/(T)`
Explanation: `DeltaS `=Entropy change
T=Temperature in K [T=`298` K under standard conditions]
`q_("surr")` = Heat absorbed by the surrounding (`q_("surr") =+286 KJ mol^(-1))`
Substitution & Calculation:
`DeltaS_("surr") =(286000)/(298) = 959.73 JK^(-1)`.
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