By Sunil Bhardwaj

2823 Views


Ethyl acetate is colorless liquid and has a characteristic sweet smell. It is used in glues, nail polish removers, decaffeinating tea and coffee.

Ethyl acetate when reacts with sodium hydroxide it produces sodium hydroxide and ethyl alcohol, the reaction is called saponification and it of second order as the rate of reaction depends on concentration of both ethyl acetate and sodium hydroxide.

Saponification of Ethyl Acetate

The reaction is carried out is a water bath to maintain the constant temperature, Solution of ethyl acetate and sodium hydroxide is mixed in a flask and the flask is kept in water bath.

As the reaction proceeds the concentrations of reactants (ethyl acetate and sodium hydroxide) decreases at the same time concentration of products (sodium acetate and ethyl alcohol increases).

To find the rate of reaction we monitor the concentration of sodium hydroxide by taking small amount of reaction mixture after a fix interval of time and titrating it with standard hydrochloric acid solution from the burette.

The reaction is $$ CH_{ 3 }COOC_{ 2 }H_{ 5 } + NaOH \longrightarrow CH_{ 3 }COONa + C_{ 2 }H_{ 5 }OH $$ So at the start the amount of acid needed for NaOH in reaction mixture will be equivalent to the initial concetration of NaOH, $$ \left\{ \text { Volume of standard acid at begining of reaction } \left( { V }_{ 0 } \right) \right\} \propto \left\{ \text { Initial concentration of NaOH } \left( a \right) \right\} $$ $$ \therefore { V }_{ 0 } \propto a \qquad ... (1)$$ After time t, x amount of NaOH gets reacted, so volume of acid will now be equivalent to the concentration of NaOH remaining in reaction mixture i.e. \(\left( a - x \right)\) . $$ \left\{ \text { Volume of standard acid at time t } \left( { V }_{ t } \right) \right\} \propto \left\{ \text { Concentration of NaOH remaining } \left( a - x \right) \right\} $$ $$ \therefore { V }_{ t } \propto \left( a - x \right) \qquad ... (2)$$ Lets put the value of a from equation (1) $$ { V }_{ t } \propto \left( { V }_{ 0 } - x \right) $$ $$ \therefore x \propto { V }_{ 0 } - { V }_{ t } \qquad ... (3)$$ So, now we have values of a and x from equation (1) and equation (3) Lets substitute these values in rate equation for second order reaction. i.e. $$ \boxed { k = \frac { 1 }{ t } \frac { x }{ a\left( a - x \right) } } $$ $$ k = \frac { 1 }{ t } \frac { \left( { V }_{ 0 } - { V }_{ t } \right) }{ { V }_{ 0 }\left[ { V }_{ 0 } - \left( { V }_{ 0 } - { V }_{ t } \right) \right] } $$ $$ k = \frac { 1 }{ t } \frac { \left( { V }_{ 0 } - { V }_{ t } \right) }{ { V }_{ 0 }\left[ { V }_{ 0 } - { V }_{ 0 } + { V }_{ t } \right] } $$ $$ \therefore k = \frac { 1 }{ t } \frac { \left( { V }_{ 0 } - { V }_{ t } \right) }{ { V }_{ 0 }\left( { V }_{ t } \right) } $$ $$ or \boxed { k = \frac { 1 }{ t } \frac { \left( { V }_{ 0 } - { V }_{ t } \right) }{ { V }_{ 0 }{ V }_{ t } } } $$ Thus from the initial and final volume of acid we can calculate rate constant of saponification of ethyl acetate.

Numerical: Starting with equal concentration of ethyl acetate and sodium hydroxide, the saponification of ethyl acetate was carried out. Same volume of the reaction mixture was withdrawn at different intervals of time and titrated with the same acid. The following data were obtained:

Time (min.):051525
Acid used (ml):1610.246.134.32
Show that the reaction is of the second order.

Solution: If the reaction is of the second order, it must obey the equation $$ \boxed { k = \frac { 1 }{ t } \frac { x }{ a\left( a - x \right) } } $$ or in this case, $$ \boxed { k = \frac { 1 }{ t } \frac { \left( { V }_{ 0 } - { V }_{ t } \right) }{ { V }_{ 0 }{ V }_{ t } } } $$ where \({ V }_{ 0 }\) is 16ml.

1) at time t = 5min volume of acid \({ V }_{ t } = 10.24\)ml $$ \therefore k = \frac { 1 }{ 5 } \times \frac { \left( 16 - 10.24 \right) }{ 16 \times 10.24 } = \frac { 5.76 }{ 819.2 } = 0.0070$$ 2) at time t = 15min volume of acid \({ V }_{ t } = 6.13\)ml $$ \therefore k = \frac { 1 }{ 15 } \times \frac { \left( 16 - 6.13 \right) }{ 16 \times 6.13 } = \frac { 9.87 }{ 1471.2 } = 0.0067$$ 3) at time t = 25min volume of acid \({ V }_{ t } = 4.32\)ml$$ \therefore k = \frac { 1 }{ 25 } \times \frac { \left( 16 - 4.32 \right) }{ 16 \times 4.32 } = \frac { 11.68 }{ 1728 } = 0.0068$$ This indicates the value of k is fairly constant. So the reaction is of the second order. And the mean rate constant is 0.0068