The enthalpies 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 $- 110.5 k J$
$C + O_{2} \to C O_{2}$ $Δ H = - 393.5 k J m o l^{-} 1$

$C O + \frac{1}{2} O_{2} \to C O_{2}$ $Δ H = - 283 k J m o l^{-} 1$

$C O_{2} \to C O + \frac{1}{2} O_{2}$ $Δ H = 283 k J m o l^{-} 1$

After solving the 3 equations we get,
$C$ + $\frac{1}{2}$ $O_{2} \to C O$
$Δ H = - 393.5 + 283$
$= - 110.5 k J m o l^{- 1}$

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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

Q. The enthalpy of combustion of carbon and carbon monoxide are

- 393.5

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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:

Q. The enthalpies of combustion of carbon and carbon monoxide are

393.5

and

- 283

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The heats of combustion of carbon and carbon monoxide are -393.5 and -283.5 kJ $m o l^{- 1}$ , respectively. The heat of formation (in kJ) of carbon monoxide per mole is:

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The enthalpies of combustion of carbon and carbon monoxide are −393.5 and −283 kJ mol respectively. The enthalpy of formation of carbon monoxide per mole is:

The correct option is D −110.5kJC+O2→CO2 ΔH=−393.5kJmol−1CO+12O2→CO2 ΔH=−283kJmol−1CO2→CO+12O2 ΔH=283kJmol−1After solving the 3 equations we get,C + 12 O2→CO ΔH=−393.5+283 =−110.5 kJmol−1

The enthalpies 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: