The standard enthalpy of formation of octane C 8 H 18 is -250 kJ - mol- Calculate the enthalpy of combustion of C 8 H 18- Given - enthalpy of formation of CO 2 g and H 2 O l are -394 kJ - mol and -286 kJ - mol respectively-A- -5726 kJ - molB- -5476 kJ - molC- -5200 kJ - molD- -5310 kJ - mol

The correct option is B 5476 kJ/molC_8H_18 (g) + 252 O_2 (g) → 8CO_2(g) + 9H_2O(l) Δ_rH^∘ = 8 × ( 394 ) + 9 × ( 286) ( 250) = 5476 kJ/mol ...

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Standard XII

Chemistry

Enthalpy of Combustion

The standard ...

Question

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:

- 5200 k J / m o l

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- 5726 k J / m o l

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- 5476 k J / m o l

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- 5310 k J / m o l

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Solution

The correct option is C $- 5476 k J / m o l$
C₈H₁₈(g)+252O₂(g)→8CO₂(g)+9H₂O(l)ΔrH∘=8×(−394)+9×(−286)−(−250)=−5476 kJ/mol

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

The correct option is C −5476 kJ/molC8H18(g)+252O2(g)→8CO2(g)+9H2O(l)ΔrH∘=8×(−394)+9×(−286)−(−250)=−5476 kJ/mol

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:

The correct option is C $- 5476 k J / m o l$
C₈H₁₈(g)+252O₂(g)→8CO₂(g)+9H₂O(l)ΔrH∘=8×(−394)+9×(−286)−(−250)=−5476 kJ/mol