Chemical Thermodynamics Question 479
Question: If $ \Delta H_{f}^{o} $ for $ H_2O_2 $ and $ H_2O $ are $ -188,kJ/mole $ and $ -286,kJ/mole $ . What will be the enthalpy change of the reaction $ 2H_2O_2(l)\to 2H_2O(l)+O_2(g) $ [MP PMT 1992]
Options:
A) $ -196,kJ/mole $
B) $ 146,kJ/mole $
C) $ -494,kJ/mole $
D) $ -98,kJ/mole $
Show Answer
Answer:
Correct Answer: A
Solution:
$ 2H_2+O_2\to 2H_2O $
$ \Delta H_{f}^{o}=-285.8,kJ/mole $ -.(i) $ H_2+\frac{1}{2}O_2\to H_2O $
$ \Delta H_{f}^{o}=-286,kJ/mole $ -.(ii) eq. (i) - eq. (ii) × 2 gives the required result.
$ = -196,kJ/mole $
BETA
