Chemistry, asked by ramvicky1978, 1 year ago

At 298 K the standard enthalpies of formation of
H2O(l) and H2O2(l) are -286.0 kJ mol-1 and
-188.0 kJ mol-'. The enthalpy change for reaction
2H,02(1)→ 2H,0(1) + O2(g) will be
(1) - 948 kJ/mol
(2) - 196 kJ mol-1
(3) + 196 kJ mol-1
4)] + 948 kJ mol-1​

Answers

Answered by pranjaygupta
25

Answer:

2) -196KJ/ mol

Explanation:

∆Hf products - ∆Hf reactants

for elements in standard state ∆Hf is zero for O2(g) it's zero

2(-286)-2(-188)

-196KJ/mol

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