Question

Calculate the entropy change in surroundings when 36 g of water is formed under standard conditions. AH" of water-286kl/mole​

Answer 1

Answer:

3x+5y-12-0

now by comparing it to ax+by+c=0

we get a=3 b=5 c=-12

Answer 2

Answer:

1919.4 J mol-1K-1

Explanation:

It is given that 286 kJ mol-1of heat is evolved on the formation of 1 mol of H2O(l). So for 2 moles of water it is 2 x 286 = 572 kJ mol-1. Thus, an equal amount of heat will be absorbed by the surroundings.

qsurr = +572 kJ mol-1

Entropy change (ΔSsurr) for the surroundings =  qsurr / T

= 572 kJ mol-1  / 298k

therefore  ΔSsurr = 1919.4 J mol-1K-1