If Δ Hfo for H2O2 and H2O are -188 kJ/mol and -286 kJ/mol, what will be the enthalpy change of the reaction: 2H2O2(l) arrow 2H2O(l)+O2(.g.)

Question

If Δ Hfo for H2O2 and H2O are -188 kJ/mol and -286 kJ/mol, what will be the enthalpy change of the reaction: 2H2O2(l) arrow 2H2O(l)+O2(.g.) (A) -196 kJ (B) -494 kJ (C) 146 kJ (D) -98 kJ

Tardigrade Admin · Accepted Answer

Given reaction is 2H2O2(l) arrow 2H2O(l)+O2(.g.) textΔH textReaction texto = textΔH textf texto ( textProducts) - textΔH textf texto ( textReactants) Enthalpy change =2× (- 286)-(2 × - 188) =-196 k J

Q. If $Δ H_{f}^{o}$ 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:

$2 H_{2} O_{2} (l) \to 2 H_{2} O (l) + O_{2} (g)$

-196 kJ

-494 kJ

146 kJ

-98 kJ

Given reaction is $2 H_{2} O_{2} (l) \to 2 H_{2} O (l) + O_{2} (g)$

$ΔH_{Reaction}^{o} = ΔH_{f}^{o} (Products) - ΔH_{f}^{o} (Reactants)$

Enthalpy change $= 2 \times (- 286) - (2 \times - 188)$

$= - 196 k J$

Q. If ΔHfo​ for H2​O2​ and H2​O are -188 kJ/mol and -286 kJ/mol, what will be the enthalpy change of the reaction: 2H2​O2​(l)→2H2​O(l)+O2​(g)