Question

Estimate the standard internal energy of formation of liquid methyl acetate (methyl ethanoate, ch3cooch3) at 298 k from its standard enthalpy of formation, which is –442 kj mol-1.

Answer 1

`the reaction for formation of methyl acetate is\\</p><p>CH_(3)OH+CH_(3)COOH----------->CH_(3)COOCH_(3)+H_(2)O\\</p><p>standard internal energy \\</p><p>\delta u=\delta H -\delta(pv)=\delta H -\delta n* RT\\</p><p>where \delta n=change in number of moles=0\\</p><p>\delta u=\delta H\\</p><p>=-442kj/mol`

Answer 2

Answer : The standard internal energy of formation of liquid methyl acetate is, -432.1 kJ/mol

Explanation :

The balanced reaction of formation of liquid methyl acetate will be:

`3C(s)+3H_2(g)+O_2(g)\rightarrow CH_3COOCH_3(l)`

Formula used :

`\Delta H=\Delta U+\Delta n_gRT`

or,

`\Delta U=\Delta H-\Delta n_gRT`

where,

`\Delta H` = change in enthalpy =
`-442kJ/mol=-442000J/mol`

`\Delta U` = change in internal energy = ?

`\Delta n_g` = change in moles = 0 - 4 = -4 (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = 298 K

Now put all the given values in the above formula, we get:

`\Delta U=\Delta H-\Delta n_gRT`

`\Delta U=(-442000J/mol)-[(-4mol)* 8.314J/mol.K* 298K`

`\Delta U=-432089.712J/mol=-432.1kJ/mol`

Therefore, the standard internal energy of formation of liquid methyl acetate is, -432.1 kJ/mol