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 $ .



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