Chemical Thermodynamics Question 174

Question: If ΔHfo for H2O2 and H2O are −188,kJ/mole and −286,kJ/mole . What will be the enthalpy change of the reaction 2H2O2(l)→2H2O(l)+O2(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:

H2+O2→H2O2

ΔHfo=−188,kJ/mole -.(i) H2+12O2→H2O

ΔHfo=−286,kJ/mole -.(ii) eq. (i) - eq. (ii) × 2 gives the required result.

=−196,kJ/mole

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