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Thermodynamics

Question
CBSEENCH11006255

Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ∆fH0 = –286 kJ mol–1.

Solution

The thermochemical equation is,
space straight H subscript 2 left parenthesis straight g right parenthesis space plus space 1 half straight O subscript 2 space space space rightwards arrow space space straight H subscript 2 straight O left parenthesis l right parenthesis semicolon space triangle subscript straight f straight H to the power of 0 space equals space minus 286 space kJ space mol to the power of negative 1 end exponent
From the above equation, it is clear that when 1 mol of H2O(l) is formed, 286 kJ of heat is released. The same amount of heat is absorbed by the surroundings.
straight i. straight e. space space space space straight q subscript surr space equals space plus 286 space kJ space mol to the power of negative 1 end exponent
We space know comma space
increment straight S space equals space straight q subscript surr over straight T space equals space fraction numerator 286 space kJ space mol to the power of negative 1 end exponent over denominator 298 space straight K end fraction
space space space space equals space 0.9597 space kJ space straight K to the power of negative 1 end exponent mol to the power of negative 1 end exponent
space space space space equals space 959.7 space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent

space space