Question

What carpet Burns in a deficiency of O2 a mixture of CO and CO2 forms.Carbon Burns in excess O2 to form only CO2 and CO Burns in excess O2 to form only CO2. Calculate ΔH for C(graphite +1/2O2) →CO(g).

Answer 1

Step-by-step explanation:

From the combustion of carbon, the reactions occurring in limited oxygen conditions are:

`C(graphite) + (1)/(2)O_(2(g)) \to CO_((g))`

`C(graphite) + O_(2(g)) \to CO_(2(g))`

If it occurs in excess, then any leftover CO changes to CO2. i.e.

`C(graphite) + O_(2(g)) \to CO_(2(g))` ---- (1)

`CO_((g)) + (1)/(2)O_((g)) \to CO_(2(g))` ----- (2)

From (1), the enthalpy change is:

`\Delta H_(rxn1) = \Delta H^0_(fCO_2(g)) - ( \Delta H^0_(f C(graphite))+ \Delta H^0_(fCO_2(g))`

`\Delta H_(rxn1) =-393.5 \ kJ/mol -(0+0)`

`\Delta H_(rxn1) =-393.5 \ kJ/mol`

From (2), the enthalpy change is:

`\Delta_(rxn2) = \Delta H^0_(fCO_2(g)) - ( \Delta H^0_(fCO(g)) + (1)/(2) \Delta H^0_(fO_2(g)))`

`\Delta_(rxn2) = -393.5 \ kJ/mol -(-110.5 + (1)/(2)(0))`

`\Delta_(rxn2) = -283.0 \ kJ/mol`

Subtracting (2) from (1), we get:

`C(graphite) + O_(2(g)) \to CO_(2(g)) \ \ \ \Delta H_(rxn) = -393.5 \ kJ/mol}`

`CO_((g)) + (1)/(2) O_2(g) \to CO_(2(g))} \ \ \ \Delta H _(rxn2) = -283.0 \ kJ/mol`

`C(graphite) + O_(2(g)) \to CO (g) + (1)/(2)O_(2(g)) \ \ \ \Delta H_(rxn) = -110.5 \ kJ/mol`

`C(graphite) + (1)/(2) O_(2(g)) \to CO (g) \ \ \ \Delta H_(rxn) = -110.5 \ kJ/mol`

The enthalpy change ΔH of the reaction = -110.5 kJ/mol