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26/09/2021 11:04 am
Topic starter
Given :
(I) H2(g) + 1/2O2(g) → H2O(l);
ΔH°298K = -285.9kJ mol-1
(II) H2(g) + 1/2O2(g) → H2O(l);
ΔH°298K = -241.8kJ mol-1
The molar enthalpy of vapourisation of water will be :
(a) 241.8 kJ mol-1
(b) 22.0 kJ mol-1
(c) 44.1 kJ mol-1
(d) 527.7 kJ mol-1
1 Answer
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26/09/2021 11:09 am
Correct answer: (c) 44.1 kJ mol-1
Explanation:
Given
H2(g) + 1/2O2(g) → H2O(l);
ΔH° = -285.9kJ mol-1 ..... (1)
H2(g) + 1/2O2(g) → H2O(g);
ΔH° = -241.8kJ mol-1 ......(2)
We have to calculate
H2O(l) → H2O(g); ΔH° = ?
On subtracting eqn. (2) from eqn. (1) we get
H2O(l) → H2O(g);
ΔH° = -241.8 - (-285.9)
= 44.1 kJ mol-1