Change in enthalpy for reaction 2H2O2(l) - 2H2O(l) + O2(g) If heat of formation of H2O2(l) and H2O(l) are - 188 - 286 KJ/mol respectively: -

Question

Change in enthalpy for reaction 2H2O2(l) -> 2H2O(l) + O2(g) If heat of formation of H2O2(l) and H2O(l) are - 188 - 286 KJ/mol respectively: - (A) - 196 KJ/mol (B) + 196 KJ/mol (C) + 948 KJ/mol (D) - 948 KJ/mol

Tardigrade Admin · Accepted Answer

2 H 2 O 2(l) arrow 2 H 2 O 2(l)+ O 2(g) Δ H=? Δ H=[2 × Δ H f. of H 2 O (l)+(Δ H f. of ..O 2)] -(2 × Δ H f. of ..H 2 O 2(l))] =[(2 ×-286)+(0)-(2 ×-188)] =[-572+376] =-196 kJ / mol

Q. Change in enthalpy for reaction $2 H X_{2} O X_{2} (l) 2 H X_{2} O (l) + O X_{2} (g)$
If heat of formation of $H X_{2} O X_{2} (l)$ and $H X_{2} O (l)$ are - 188 & - 286 KJ/mol respectively : -

- 196 KJ/mol

+ 196 KJ/mol

+ 948 KJ/mol

- 948 KJ/mol

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

Q. Change in enthalpy for reaction 2HX2​OX2​(l)​2HX2​O(l)+OX2​(g) If heat of formation of HX2​OX2​(l) and HX2​O(l) are - 188 & - 286 KJ/mol respectively : -