1.

Calculate thework done during combustion of 0.138 kg of ethanol, C_(2)H_(5)OH_(1) at 300 K. Given: R= 8.314 JK^(-1) mol^(-1) molar mass of ethanol = 46 gm mol^(-1)

Answer»

`-7482` J
7482 J
`-2494` J
2494 J

Solution :`underset(0.138 kg) underset(darr)(C_(2)H_(5)OH) +underset(3)underset(darr)(3O_(2)) to underset(2)underset(darr)(2CO_(2)) +underset(3)underset(darr)(3H_(2)O)`
Hence Mole `=138/46 =3 `moles of `C_(2)H_(5)OH`.
`3C_(2)H_(5)OH(l) + 9O_(g) to 6CO_(2)(g) + 9H_(2)O (u)`
`deltan = 6-9 =-3`
WORK =`-deltanRT`
`=-(-3) xx 8.314 xx 300`
`=-7482` J


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