Question

Calculate the Standard internal energy of formation of liquid methyl acetata $\left(C{H}_{3}COOC{H}_3\right)$ from standard enthalpy of formation, which is $-442.9Kkmol{e}^{-1}$ at ${25}^0$C.

• A. $-433$
• B. $+433$
• C. $-452.82$
• D. $452.82$

Answer 1

The correct option is C $-452.82$

Solution:- (C) $-452.82$

Formation of liquid methyl acetate-

$3C_{\left(s\right)}+3{H_2}_{\left(g\right)}+{O_2}_{\left(g\right)}⟶{C{H}_{3}COOC{H}_3}_{\left(l\right)}$

$\Delta n_g=0-(1+3)=-4$

Given:- $\Delta H=-442.9kJ/mol$

$T=25℃=(25+273)K=298K$

As we know that,

$\Delta H=\Delta U+\Delta n_{g}RT$

$\therefore \Delta U=\Delta H-\Delta n_{g}RT$

$\Rightarrow \Delta U=-442.9+(-4\times 8.314\times {10}^{-3}\times 298)$

$\Rightarrow \Delta U=-442.9-9.92=-452.82kJ/mol$

Hence the standard internal energy of formation of liquid methyl acetate is $-452.82kJ/mol$.