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  • Stuck here, help me understand: - Chemical Thermodynamics - JEE Main

(\Delta H-\Delta U) for the formation of carbon monoxide (CO) from its elements at 298 K is

( R = 8.314 JK-1 mol-1  )

  • Option 1)

    –1238.78 J mol-1

  • Option 2)

    1238.78 J mol-1

  • Option 3)

    –2477.57 J mol-1

  • Option 4)

    2477.57 J mol-1

 

Answers (1)

best_answer

As we learnt in

Enthalpy of combustion -

Amount of Enthalpy change when 1 mole of substance is completly burn in excess of Oxygen

- wherein

CO_{(g)}+\frac{1}{2}O_{2}(g)\rightarrow CO_{2(g)}+283\ KJ

\Delta H_{comb} \: of\: CO= 283\, KJ

 

 C(S)+\frac{1}{2}O_{2}(g)\rightarrow CO(g)

\Delta n_{g}=1-\frac{1}{2}=\frac{1}{2}

\Delta H-\Delta U=\Delta n_{g}RT=\frac{1}{2}\:J/mol\times8.314JK^{-1}mol^{-1}\times298

= 1238.78 J/mol


Option 1)

–1238.78 J mol-1

This option is incorrect.

Option 2)

1238.78 J mol-1

This option is correct.

Option 3)

–2477.57 J mol-1

This option is incorrect.

Option 4)

2477.57 J mol-1

This option is incorrect.

Posted by

divya.saini

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