The enthalpy of combustion of carbon and carbon monoxide are $- 393.5$ and $- 283 k J / m o l$ respectively. The enthalpy of formation of carbon monoxide per mole is:

110.5 k J

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676.5 k J

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- 676.5 k J

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- 110.5 k J

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Solution

The correct option is D $- 110.5 k J$
I: $C (s) + O_{2} (g) \to C O_{2}; Δ H_{1} = - 393.5 k J / m o l$

II: $C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g); Δ H_{2} = - 283.0 k J / m o l$

On substracting II from I we get the formation reaction of carbon monoxide.

III: $C (s) + \frac{1}{2} O_{2} (g) \to C O (g); Δ H = Δ H_{1} - Δ H_{2} = - 110.5 k J / m o l$

This equation III represents the formation of $1$ mole of $C O$ and thus enthalpy of formation of $C O$ (g) is $- 110.5 k J / m o l$ .

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The enthalpies of combustion of carbon and carbon monoxide are $- 393.5$ and $- 283 k J m o l^{- 1}$ respectively. The enthalpy of formation of carbon monoxide per mole is

The enthalpies of combustion of carbon and carbon monoxide are $- 393.5$ and $- 283$ kJ mol $^{- 1}$ respectively. The enthalpy of formation of carbon monoxide per mole is: