Chemical Thermodynamics Question 252
Question: 1 gram equivalent of $ H_2SO_4 $ is treated with 112 g of $ KOH $ for complete neutralization. Which of the following statements is correct-
Options:
A) 13.7 kcal of heat is evolved with the formation of 87 g of $ K_2SO_4 $ , leaving no KOH.
B) 27.4 kcal of heat is evolved with the formation of 87 g of $ K_2SO_4 $ , leaving 4 gram equivalent of KOH.
C) 15.7 kcal of heat is evolved with the formation of 1 gram equivalent of $ K_2SO_4 $ , leaving 56 g of KOH.
D) 13.7 kcal of heat is evolved with the formation of 87g of $ K_2SO_4 $ , leaving 1 gram equivalent of KOH.
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Answer:
Correct Answer: D
Solution:
- $ \underset{49}{\mathop{\underset{98}{\mathop{H_2SO_4}},}},+\underset{56}{\mathop{\underset{112}{\mathop{2KOH}},}},\to \underset{87}{\mathop{\underset{174}{\mathop{K_2SO_4}},}},+\underset{1,mole}{\mathop{\underset{2,mole}{\mathop{2H_2O}},}}, $ 13.7 kcal is the heat evolved when 1 gev of strong acid is neutralised by 1 gev of strong base.
BETA
