Question

Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions at 298 K. Given ΔrH0 = - 286 kJ mol1.

Solution
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H2(g)+12O2(g)H2O(l)ΔfH0=286KJ/mol
From the above equation,

At 298K, when 1 mole of H2O(l) is formed, 286KJ of heat is released. The same amount of heat is absorbed by the surroundings.

qsurr.=+286KJ/mol;T=298K

As we know that,

ΔSsurr.=qsurr.T

ΔSsurr.=286298=0.96KJ/molK

Hence the entropy change in surroundings will be 0.96KJ/mol.

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