The change in entropy of surroundings, when 1 mole of liquid H2O is formed from its constituent
elements in their standard state at 298 K, is equal to X J/K. If AHºformation for liquid water = –286.08
X Х
kJ/mol then report the value of
100
Answers
Answered by
0
Explanation:
H
2
(g)
+
2
1
O
2
(g)
⟶H
2
O
(l)
Δ
f
H
0
=−286KJ/mol
From the above equation,
At 298K, when 1 mole of H
2
O
(l)
is formed, 286KJ of heat is released. The same amount of heat is absorbed by the surroundings.
∴q
surr.
=+286KJ/mol;T=298K
As we know that,
ΔS
surr.
=
T
q
surr.
∴ΔS
surr.
=
298
286
=0.96KJ/mol−K
Hence the entropy change in surroundings will be 0.96KJ/mol.
Similar questions