Chemical Thermodynamics Question 281
Question: Heat of formation of $ H_2O $ is $ -188,kJ/mole $ and $ H_2O_2 $ is $ -286,kJ/mole. $ The enthalpy change for the reaction $ 2H_2O_2,\to ,2H_2O+O_2 $ is [BHU 2001]
Options:
A) 196 kJ
B) - 196 kJ
C) 984 kJ
D) - 984 kJ
Show Answer
Answer:
Correct Answer: A
Solution:
$ H_2+\frac{1}{2}O_2\to H_2O,\Delta H=-188,kJ/mole $ ..-..(i) $ H_2+O_2\to H_2O_2;,\Delta H=-286,kJ/mole $ ..-.(ii) By 2 × (i) and (ii) $ 2H_2+O_2\to 2H_2O,;\Delta H=-376,kJ/mole $ -..(iii) $ 2H_2+2O_2\to 2H_2O_2,\Delta H=-572,kJ/mole $ -..(iv) By (iii) - (iv) $ 2H_2O_2\to 2H_2O+O_2,\Delta H=+196,kJ $ .
