# 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 mol−1.

#### H2(g)+12O2(g)⟶H2O(l)ΔfH0=−286KJ/molFrom 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/mol−K

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