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(g…

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asked Feb 25, 2022 112k views

1 Answer

Marcin Kunert · Answer 1 · 2022-03-02T23:53:11+0000

Answer:

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