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The enthalpy of combustion of solid carbon to form carbon dioxide is -393.7 kJ/mol carbon, and the enthalpy of combustion of carbon monoxide to form carbon dioxide is -283.3 kj/mol CO. Use these data to calculate ∆H for the reaction.

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4 answers

  1. If the stoichiometric coefficients are multiplied by some number, then the value for ∆H (in kJ/mol) is also multiplied by that number.

    We got earlier the ∆H formation for CO with fraction stoichiometric coefficients:
    C + 1/2 O2 ---> CO ; ∆H = -110.4 kJ/mol

    And we need ∆H for 2C(s) + O2(g) → 2CO(g). We just multiply everything by 2:

    2 * [ C + 1/2 O2 ---> CO ; ∆H = -110.4 kJ/mol ]
    = 2C(s) + O2(g) → 2CO(g) ; ∆H = -220.8 kJ

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  2. What reaction?

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  3. ∆H for what reaction...?

    Write the balanced reactions:
    C + O2 ---> CO2 ; ∆H = -393.7 kJ/mol C

    CO + 1/2 O2 ---> CO2 ; ∆H = -283.3 kJ/mol CO

    If the problem asks for ∆H of formation for CO, we can make the second reaction become:
    CO2 ---> CO + 1/2 O2 ; ∆H = +283.3 kJ/mol CO

    And add this to the first equation:
    CO2 ---> CO + 1/2 O2 ; ∆H = +283.3 kJ/mol
    C + O2 ---> CO2 ; ∆H = -393.7 kJ/mol
    --------------------------------------------------
    C + 1/2 O2 ---> CO ; ∆H = -110.4 kJ/mol

    hope this helps? `u`

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  4. I sincerely apologize. Here's the reaction:

    2C(s) + O2(g)→2CO(g)

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