Question

The equilibrium constant of ester formation of propionic acid with ethyl alcohol is \[7.36\] ${50^\circ }C$. Calculate the weight of ethyl propionate in gram existing in an equilibrium mixture when $0.5$ mole of propionic acid is heated with $0.5$ mole of ethyl alcohol at ${50^\circ }C$.

Answer 1

Hint:We will use the formula of equilibrium constant. The equilibrium constant is the ratio of the concentration of products to the concentration of reactants with the concentration raised to the appropriate stoichiometric coefficient. It is written as-
$ \Rightarrow $${K_{eq}}$=$\dfrac{{\left[ {{\text{Product}}} \right]}}{{\left[ {{\text{Reactant}}} \right]}}$ where $\left[ {} \right]$ is a symbol for concentration
Then use the formula which gives relation between number of moles and mass of substance as-
$ \Rightarrow $ n=$\dfrac{{\text{w}}}{{\text{M}}}$ where n is number of moles, M is molar mass and w is given mass of the substance.

Complete step-by-step answer:Here, it is given that equilibrium constant ${K_{eq}}$of ester formation of propionic acid $\left( {{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{COOH}}} \right)$ with ethyl alcohol $\left( {{{\text{C}}_2}{{\text{H}}_5}{\text{OH}}} \right)$ is \[7.36\]at${50^\circ }C$.
 Also given, propionic acid of concentration $0.5$ mole is mixed with ethyl alcohol of concentration $0.5$ mole. When propionic acid is mixed with ethyl alcohol, ethyl propionate is formed and the reaction is given as-
$\mathop {{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{COOH}}}\limits_{{\text{propionic acid}}} {\text{ + }}\mathop {{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{OH}}}\limits_{{\text{ethyl alcohol}}} \xrightarrow{{{{\text{H}}^ + }}}\mathop {{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{COOC}}{{\text{H}}_2}{\text{C}}{{\text{H}}_3}}\limits_{{\text{ethyl propionate}}} + {{\text{H}}_2}{\text{O}}$
We have to calculate the weight of ethyl propionate in gram existing in the equilibrium mixture.
Here, Initial concentration of propionic acid=$0.5$ mole
And initial concentration of ethyl alcohol=$0.5$ mole
Initial concentration of products= $0$mole
Let the number of moles of reactant decomposed or the number of moles of product formed be x. Then concentration at equilibrium will be-
Propionic acid=$\left( {0.5 - x} \right)$ mole
Ethyl alcohol=$\left( {0.5 - x} \right)$ mole
Ethyl Propionate=x mole
Water=x mole
Now, we know that equilibrium constant can be written the ratio of the concentration of products to the concentration of reactants with the concentration raised to the appropriate stoichiometric coefficient.
$ \Rightarrow $${K_{eq}}$=$\dfrac{{\left[ {{\text{Product}}} \right]}}{{\left[ {{\text{Reactant}}} \right]}}$ where $\left[ {} \right]$ is a symbol for concentration
Then for the above reaction, equilibrium constant is-
$ \Rightarrow $${K_{eq}}$=$\dfrac{{\left[ {{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{COOC}}{{\text{H}}_2}{\text{C}}{{\text{H}}_3}} \right]\left[ {{{\text{H}}_2}{\text{O}}} \right]}}{{\left[ {{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{COOH}}} \right]\left[ {{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{OH}}} \right]}}$
On putting the given values, we get-
$ \Rightarrow $$7.36 = \dfrac{{x \times x}}{{\left( {0.5 - x} \right) \times \left( {0.5 - x} \right)}}$
_{On solving, we get-}
$ \Rightarrow 7.36 = \dfrac{{{x^2}}}{{{{\left( {0.5 - x} \right)}^2}}}$
On removing the square-root from right side, we get-
$ \Rightarrow \sqrt {7.36} = \dfrac{x}{{\left( {0.5 - x} \right)}}$
On rearranging, we get-
$ \Rightarrow \left( {0.5 - x} \right)\sqrt {7.36} = x$
On simplifying, we get-
$ \Rightarrow 0.5\sqrt {7.36} - \sqrt {7.36} x = x$
On simplifying again, we get-
$ \Rightarrow 0.5\sqrt {7.36} = \left( {1 + \sqrt {7.36} } \right)x$
On arranging, we get-
$ \Rightarrow x = \dfrac{{0.5\sqrt {7.36} }}{{1 + \sqrt {7.36} }}$
On solving, we get-
$ \Rightarrow x = 0.3653$
Now, we know that number of moles is related to mass of substance as-
$ \Rightarrow $ n=$\dfrac{{\text{w}}}{{\text{M}}}$ where n is number of moles, M is molar mass and w is given mass of the substance
Here molar mass of ethyl propionate=$\left( {12 \times 5} \right) + \left( {1 \times 10} \right) + \left( {16 \times 2} \right) = 102$ g/mol
$ \Rightarrow 0.3653 = \dfrac{{\text{w}}}{{102}}$
On solving, we get-Weight of ethyl propionate=$0.3653 \times 102 = 37.2606$g

Note:To calculate the molar mass of a compound, we multiply the atomic mass of the elements present in the compound by the number in which they are present individually and add them all. Also, to calculate the moles of product formed at equilibrium we subtract the number of moles of reactants decomposed from the initial concentration of the reactants.