Question

The reaction$CH_{3}COOH(l)+C_{2}H_{5}OH(l) \rightleftharpoons CH_{3}COOC_{2}C_{5}(l)+H_{2}O(l)$ was carried out at 27°C by taking one mole of each reactants. The equilibrium constant is 4. Calculate the standard free energy change for the reaction (R = 8.314 J$K^{-1}$ $mol^{-1}$).

Answer 1

The given chemical reaction is as follows:

             ${ CH }_{ 3 }COOH(l)\quad +\quad { C }_{ 2 }{ H }_{ 5 }OH(l)\quad \longrightarrow \quad { CH }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 }(l)\quad +\quad { H }_{ 2 }O(l)$

Let's write the equilibrium constant

             $K\quad =\quad \frac { [{ CH }_{ 3 }COO{ C }_{ 2 }{ H }_{ 5 }][{ H }_{ 2 }O] }{ [{ CH }_{ 3 }COOH][{ C }_{ 2 }{ H }_{ 5 }OH] } \quad =\quad 4$

              ${ \Delta G }^{ \circ  }\quad =\quad -2.303RTlogK$

              $=\quad -2.303\quad \times \quad 8.31J{ K }^{ -1 }{ mol }^{ -1 }\quad \times \quad 300\quad K\quad \times \quad log4$

              $=\quad -3458\quad J{ K }^{ -1 }{ mol }^{ -1 }$

$\Delta G^\circ\quad =\quad -3.458\quad J{ K }^{ -1 }{ mol }^{ -1 }$

Answer 2

Here is your answer mate.....☺✌☺
∆G°=-3.458J/k×mol