The heats of combustion of carbon and carbon monoxide are –393.5 and –283.5 kJ mol–1, respectively. The heat of formation (in kJ) of carbon monoxide per mole is :

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Question

The heats of combustion of carbon and carbon monoxide are –393.5 and –283.5 kJ mol^–1, respectively. The heat of formation (in kJ) of carbon monoxide per mole is :

–676.5

676.5

110.5

–110.5

$C + O_{2} \underset{}{\to} {CO}_{2}$ …………(1)

$CO + \frac{1}{2} O_{2} \underset{}{\to} {CO}_{2}$ …………(2)

$Δ$ H₁ ⇒ –393.5 KJ /Mole

$Δ$ H₂ ⇒ –283.5 KJ /mole

$C + \frac{1}{2} O_{2} \underset{}{\to} CO$

$Δ$ _fH_(CO,g)= $Δ$ H₁ – $Δ$ H₂

⇒ – 393.5 –(–283.5)

=–393.5 + 283.5

⇒ –110 KJ /Mole

The question involves calculating the heat of formation of carbon monoxide using given heats of combustion. The heat of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states.

Step 1: Write the combustion reactions and their enthalpies

Combustion of carbon (to CO₂):
$C (s) + O_{2} (g) \to C O_{2} (g); Δ H = - 393.5 kJ / mol$

Combustion of carbon monoxide (to CO₂):
$CO (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g); Δ H = - 283.5 kJ / mol$

Step 2: Write the formation reaction for CO

We want the reaction:
$C (s) + \frac{1}{2} O_{2} (g) \to CO (g); Δ H_{f} = ?$

Step 3: Use Hess's Law to find ΔH_f

Notice that the formation reaction can be obtained by subtracting the combustion of CO from the combustion of C:

Combustion of C: C(s) + O₂(g) → CO₂(g); ΔH = –393.5 kJ/mol

Reverse combustion of CO: CO₂(g) → CO(g) + ½O₂(g); ΔH = +283.5 kJ/mol (reverse sign)

Adding these: C(s) + O₂(g) + CO₂(g) → CO₂(g) + CO(g) + ½O₂(g)

Simplifying: C(s) + ½O₂(g) → CO(g)

So, ΔH_f = ΔH_combustion(C) + (–ΔH_combustion(CO)) = –393.5 + 283.5 = –110.0 kJ/mol

(Note: The calculation gives –110.0 kJ/mol, but the options include –110.5, which is close and likely the intended answer based on typical values.)

Final Answer: –110.5 kJ/mol

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