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The enthalpies of combustion of carbon and carbon monoxide are –393.5 and –283 kJ mol-1  respectively. The enthalpy of formation of carbon monoxide per mole is  

  • Option 1)

    110.5 kJ

  • Option 2)

    676.5 kJ

  • Option 3)

    –676.5 kJ

  • Option 4)

    –110.5 kJ.

 

Answers (3)

As we learnt in

Hess's Law -

The total amount of heat change in a chemical reaction is the same whether the reaction is carried  out in one or several steps by one or more methods.

- wherein

C_{(s)}+O_{2(g)}\rightarrow CO_{2(g)}

\Delta H= 393.5\, kj

C_{(s)}+\frac{1}{2}O_{2(g)}\rightarrow CO_{(g)}

\Delta H_{1}= 110.5\, kj

CO_{(g)}+\frac{1}{2}O_{2(g)}\rightarrow CO_{2(g)}

\Delta H_{2}= 283.0\, kj

\Delta H= \Delta H_{1}+\Delta H_{2}

 

 i) C + O2\rightleftharpoons CO2  \DeltaH = -393.5KJ mol-1

ii) CO + \frac{1}{2}O2\rightleftharpoons CO, \Delta H = - 283.0 kJ mol-1

operating (i) and (ii)

C +\frac{1}{2}O_{2}\rightarrow CO\Delta H=-110.5 kJ mol^{-1}


Option 1)

110.5 kJ

this is incorrect option

Option 2)

676.5 kJ

this is incorrect option

Option 3)

–676.5 kJ

this is incorrect option

Option 4)

–110.5 kJ.

this is correct option

Posted by

Sabhrant Ambastha

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Posted by

Manoj

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_110.5 kJ

 

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aadhil

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