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Standard XII

Chemistry

Enthalpy of Combustion

The standard ...

Question

The standard enthalpy of formation of octane $(C_{8} H_{18})$ is $- 250 k J / m o l$ . Calculate the enthalpy of combustion of $C_{8} H_{18}$ .
Given : enthalpy of formation of $C O_{2} (g)$ and $H_{2} O (l)$ are $- 394 k J / m o l$ and $- 286 k J / m o l$ 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 $- 5476 k J / m o l$
$C_{8} H_{18} (g) + \frac{25}{2} O_{2} (g) \to 8 C O_{2} (g) + 9 H_{2} O (l) Δ_{r} H^{\circ} = 8 \times (- 394) + 9 \times (- 286) - (- 250) = - 5476 k J / m o l$

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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 $(C_{8} H_{18})$ is $- 250 k J / m o l$ . Calculate the enthalpy of combustion of $C_{8} H_{18}$ .
Given : enthalpy of formation of $C O_{2} (g)$ and $H_{2} O (l)$ are $- 394 k J / m o l$ and $- 286 k J / m o l$ respectively:

The correct option is C $- 5476 k J / m o l$
$C_{8} H_{18} (g) + \frac{25}{2} O_{2} (g) \to 8 C O_{2} (g) + 9 H_{2} O (l) Δ_{r} H^{\circ} = 8 \times (- 394) + 9 \times (- 286) - (- 250) = - 5476 k J / m o l$