Chemical Thermodynamics Question 214

Question: H2+12O2H2O;,ΔH=68.39,kcal

K+H2O+ Water KOH(aq)+12H2;,ΔH=48,kcal

KOH+ Water KOH(aq);,ΔH=14,kcal The heat of formation of KOH is (in kcal) [CPMT 1988]

Options:

A) 68.39+4814

B) 68.3948+14

C) 68.3948+14

D) 68.39 + 48 + 14

Show Answer

Answer:

Correct Answer: B

Solution:

Aim: K(S)+12O2(g)+12H2(g)KOH(S) eq. (ii) + eq. (i) - eq. (iii) gives ΔH=48+(68.39)(14)

=68.3948+14 .



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