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 $ .



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