The standard enthalpy of formation of octane (C₈H₁₈) is −250 kJ/mol. Calculate the enthalpy of combustion of C₈H_18.
Given : enthalpy of formation of CO₂(g) and H₂O(l) are −394 kJ/mol and −286 kJ/mol respectively:

Q. The standard enthalpies of combustion of

C_{6} H_{6} (l),

C_{(g r a p h i t e)}

and

H_{2} (g)

are respectively

- 3270

k J

{m o l}^{- 1},

- 394

k J

{m o l}^{- 1}

and

- 286

k J

{m o l}^{- 1}

. What is the standard enthalpy of formation of

C_{6} H_{6} (l)

k J

{m o l}^{- 1}

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If ΔH0f for H2O2 and H2O are -188 kJ/mol and -286 kJ/mol. What will be the enthalpy change of the reaction: 2H2O2(l)→2H2O(l)+O2(g)?

The correct option is A -196 kJ/mol2H2O2(l)→2H2O(l)+O2(g)ΔH=2×ΔfH(H2O(l))+ΔfH(O2(g))−2×ΔfH(H2O2(l))=2×(−286)+0−2×(−188)=2(−286+188)=2×(−98)=−196 kJ/mol

If $Δ H_{f}^{0}$ 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: $2 H_{2} O_{2} (l) \to 2 H_{2} O (l) + O_{2} (g)$ ?

The correct option is A -196 kJ/mol
$2 H_{2} O_{2} (l) \to 2 H_{2} O (l) + O_{2} (g)$
$Δ H = 2 \times Δ_{f} H_{(H_{2} O (l))} + Δ_{f} H_{(O_{2} (g))} - 2 \times Δ_{f} H_{(H_{2} O_{2} (l))}$
$= 2 \times (- 286) + 0 - 2 \times (- 188)$
$= 2 (- 286 + 188)$
$= 2 \times (- 98)$
$= - 196$ kJ/mol