Answer:
Explanation:
From the combustion of carbon, the reactions occurring in limited oxygen conditions are:
[tex]C(graphite) + \dfrac{1}{2}O_{2(g)} \to CO_{(g)}[/tex]
[tex]C(graphite) + O_{2(g)} \to CO_{2(g)}[/tex]
If it occurs in excess, then any leftover CO changes to CO2. i.e.
[tex]C(graphite) + O_{2(g)} \to CO_{2(g)}[/tex] ---- (1)
[tex]CO_{(g)} + \dfrac{1}{2}O_{(g)} \to CO_{2(g)}[/tex] ----- (2)
From (1), the enthalpy change is:
[tex]\Delta H_{rxn1} = \Delta H^0_{fCO_2(g)} - ( \Delta H^0_{f C(graphite)}+ \Delta H^0_{fCO_2(g)}[/tex]
[tex]\Delta H_{rxn1} =-393.5 \ kJ/mol -(0+0)[/tex]
[tex]\Delta H_{rxn1} =-393.5 \ kJ/mol[/tex]
From (2), the enthalpy change is:
[tex]\Delta_{rxn2} = \Delta H^0_{fCO_2(g)} - ( \Delta H^0_{fCO(g)} + \dfrac{1}{2} \Delta H^0_{fO_2(g)})[/tex]
[tex]\Delta_{rxn2} = -393.5 \ kJ/mol -(-110.5 + \dfrac{1}{2}(0))[/tex]
[tex]\Delta_{rxn2} = -283.0 \ kJ/mol[/tex]
Subtracting (2) from (1), we get:
[tex]C(graphite) + O_{2(g)} \to CO_{2(g)} \ \ \ \Delta H_{rxn} = -393.5 \ kJ/mol}[/tex]
[tex]CO_{(g)} + \dfrac{1}{2} O_2(g) \to CO_{2(g)}} \ \ \ \Delta H _{rxn2} = -283.0 \ kJ/mol[/tex]
[tex]C(graphite) + O_{2(g)} \to CO (g) + \dfrac{1}{2}O_{2(g)} \ \ \ \Delta H_{rxn} = -110.5 \ kJ/mol[/tex]
[tex]C(graphite) + \dfrac{1}{2} O_{2(g)} \to CO (g) \ \ \ \Delta H_{rxn} = -110.5 \ kJ/mol[/tex]
The enthalpy change ΔH of the reaction = -110.5 kJ/mol
