If ∆ H f o for H 2 O 2 and H 2 O are -188 kJ/mol and -286 kJ/mol, what will be the enthalpy change of the reaction: 2 H 2 O 2 l → 2 H 2 O l + O 2 ( g )

Question

If ∆ H f o for H 2 O 2 and H 2 O are -188 kJ/mol and -286 kJ/mol, what will be the enthalpy change of the reaction: 2 H 2 O 2 l → 2 H 2 O l + O 2 ( g ), -196 kJ, -494 kJ, 146 kJ, -98 kJ

MARKS App · Accepted Answer

Given reaction is 2 H 2 O 2 l → 2 H 2 O l + O 2 ( g ) ΔH Reaction o = ΔH f o Products − ΔH f o Reactants Enthalpy change = 2 × - 286 - 2 × - 188 = - 196 k J

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