Given :
Number of moles of ethanol,
(C2H5OH) = n = 1 mol
ΔfH0(CH3OH) = -238.9 kJ mol-1
= -238.9 × 103J mol-1
Temperature = T = 298 K
ΔS = ?
ΔSsurr = ?
Since ΔfH0 is negative,
The reaction for the formation of one mole of C2H5OH is exothermic.
As heat is released to the surroundings,
ΔH0surr = + 238.9 kJ mol-1
∴ ΔSsurr = \(\frac{ΔH^0_{surr}}{T}\) = \(\frac{+238.9\times 10^3}{298}\)
= +801.7JK-1
Thus,
Entropy of the surroundings increases.
∴ ΔSsurr = +801.7JK-1