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For liquid ethanol at 25^(@)C,DeltaH(f)^...

For liquid ethanol at `25^(@)C,DeltaH_(f)^(0)=-277.69kJ*mol^(-1)`. Represent the thermochemical equation showing the formation reaction of liquid ethanol.

Text Solution

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2C(graphite)`+3H_(2)(g)+(1)/(2)O_(2)toC_(2)H_(5)OH(l),DeltaH_(f)^(0)=-277.69kJ*mol^(-1)`.
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Knowledge Check

  • Given, C(graphite)+ O_(2)(g)toCO_(2)(g),Delta_(r)H^(0)=-393.5kJ*mol^(-1) H_(2)(g)+(1)/(2)(g)toH_(2)O(l),Delta_(r)H^(0)=-285.8kJ*mol^(-1) CO_(2)(g)+2H_(2)O(l)toCH_(4)(g)+2O_(2)(g),Delta_(r)DeltaH^(0)=+890.3kJ*mol^(-1) Based on the above thermochemical equations, the value of Delta_(r)H^(0) at 298K for the reaction, C(graphite) +2H_(2)(g)toCH_(4)(g) , will be-

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    B
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  • The formation of the oxide ion O^(2-)(g) , form oxygen atom requires first an exothermic and then an endothermic step as shown below, O(g)+e^(-) to O^(-)(g),DeltaH_(f)^(0)=-141kJ*mol^(-1) O^(-)(g)+e^(-) to O^(2-)(g),DeltaH_(f)^(0)=+780kJ*mol^(-1) ltbr. Thus, process of formations of O^(2-) in gas phase is unfavourable even through O^(2-) is isoelectronic with neon. it is due to the fact that-

    A
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    B
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    C
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    A
    `-113.5`
    B
    `-12.1`
    C
    `-102.5`
    D
    `-35.7`

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