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

solution:

$the reaction for formation of methyl acetate is\\

CH_{3}OH+CH_{3}COOH----------->CH_{3}COOCH_{3}+H_{2}O\\

standard internal energy \\

\delta u=\delta H -\delta(pv)=\delta H -\delta n\times RT\\

where \delta n=change in number of moles=0\\

\delta u=\delta H\\

=-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:

$\Delta U=\Delta H-\Delta n_gRT$

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

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

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