Chemical Thermodynamics Question 214
Question: $ H_2+\frac{1}{2}O_2\to H_2O;,\Delta H=-68.39,kcal $
$ K+H_2O+ $ Water $ \to KOH(aq)+\frac{1}{2}H_2;,\Delta H=-48,kcal $
$ KOH+ $ Water $ \to KOH(aq);,\Delta H=-14,kcal $ The heat of formation of $ KOH $ is (in kcal) [CPMT 1988]
Options:
A) $ -68.39+48-14 $
B) $ -68.39-48+14 $
C) $ 68.39-48+14 $
D) 68.39 + 48 + 14
Show Answer
Answer:
Correct Answer: B
Solution:
Aim: $ {K_{(S)}}+\frac{1}{2}O_2{(g)}+\frac{1}{2}{H{2(g)}}\to KO{H_{(S)}} $ eq. (ii) + eq. (i) - eq. (iii) gives $ \Delta H=-48+(-68.39)-(-14) $
$ =-68.39-48+14 $ .
