Change in enthalpy for reaction 2H2O2(l)→ 2H2O(l) + O2(g) if heat of formation of H2O2(l) and H2O(l) are -188 and -286 kJ/mol respectively is

Question

Change in enthalpy for reaction 2H2O2(l)→ 2H2O(l) + O2(g) if heat of formation of H2O2(l) and H2O(l) are -188 and -286 kJ/mol respectively is (A) -196 kJ/mol (B) +196 kJ/mol (C) +948 kJ/mol (D) -948 kJ/mol

Tardigrade Admin · Accepted Answer

2H2O2(l)→2H2O(l)+O2 Δ H=? Δ H=[(2×Δ Hf of H2O(l)+Δ Hf of O2 )]-[(2×Δ Hf of H2O2(l))] [(2×-286)+(0)-(2×188)] =[-572+376]=-196 kJ/mol

Q. Change in enthalpy for reaction
$2 H_{2} O_{2} (l) \to 2 H_{2} O (l) + O_{2} (g)$
if heat of formation of $H_{2} O_{2} (l)$ and $H_{2} O (l)$ are $- 188$ and $- 286 k J / m o l$ respectively is

$- 196 k J / m o l$

$+ 196 k J / m o l$

$+ 948 k J / m o l$

$- 948 k J / m o l$

$2 H_{2} O_{2} (l) \to 2 H_{2} O (l) + O_{2} Δ H = ?$
$Δ H = [(2 \times Δ H_{f} o f H_{2} O (l) + Δ H_{f} o f O_{2})] - [(2 \times Δ H_{f} o f H_{2} O_{2} (l))]$
$[(2 \times - 286) + (0) - (2 \times 188)]$
$= [- 572 + 376] = - 196 k J / m o l$

Q. Change in enthalpy for reaction 2H2​O2​(l)→2H2​O(l)+O2​(g) if heat of formation of H2​O2​(l) and H2​O(l) are −188 and −286kJ/mol respectively is

−196kJ/mol

+196kJ/mol

+948kJ/mol

−948kJ/mol