Question

1.0 mole of ethyl alcohol and 1.0 mole of acetic acid are mixed.at equilibrium 0.666 moles of ester is present. The value of equilibrium constant is:-

a) 1/4
b)1/2
c) 2
d) 4

need solution for this!!​

Answer 1

★ Concept :-

Here the concept of Equilibrium has been used. We see that we are given the initial moles of ethyl alcohol and acetic acid. So firstly from the number of moles of ester at equilibrium, we can find the number of moles of water, acetic acid and ethyl alcohol. And then by using the formula of Equilibrium Constant, we can find the required value.

Let's do it !!

______________________________________

★ Formula Used :-

For a reaction,

A + B ⇌ C + D

$\;\boxed{\sf{\pink{Equilibrium\;Constant\;=\;\bf{\dfrac{[C]\:[D]}{[A]\:[B]}}}}}$

______________________________________

★ Solution :-

Given reaction ,

→ C₂H₅OH + CH₃COOH ⇌ CH₃COOC₂H₅ + H₂O

Here ,

» Initial moles of C₂H₅OH = 1 mole

» Initial moles of CH₃COOH = 1 mole

Since when the reaction is at equilibrium,

• At equilibrium, moles of C₂H₅OH = n

• Similarly, since both of them are equal initially, so at equilibrium, moles of CH₃COOH = n

• At equilibrium, moles of CH₃COOC₂H₅ = 0.666 moles = e (say)

• Since water is also formed which is also at equilibrium, so at equilibrium, moles of H₂O = 0.666 moles = e (say)

-----------------------------------------------------------

~ For the value of n ::

Since those all reactants and products are in equilibrium. So at equilibrium their number of moles will be given as,

⟩⟩ n = 1 - e

⟩⟩ n = 1 - 0.666

⟩⟩ n = 0.334

Hence,

• At equilibrium, number of moles of C₂H₅OH = 0.334

• At equilibrium, number of moles of CH₃COOH = 0.334

-----------------------------------------------------------

~ For the concentration of reactants ::

Since we see that the number of moles of reactants are equal and number of moles of products are also equal. So their molarity (concentration) per volume will be also equal. So,

$\;\bf{\rightarrow\;\;\orange{Molarity\;=\;\dfrac{Number\;of\;Moles}{Volume\;of\;Solution}}}$

• For reactants ::

By applying values in formula, we get

$\;\bf{\rightarrow\;\;Molarity\;=\;\dfrac{0.334}{V}}$

• For Products ::

By applying values, we get

$\;\bf{\rightarrow\;\;Molarity\;=\;\dfrac{0.666}{V}}$

-----------------------------------------------------------

~ For Equilibrium Constant ::

In finding equilibrium Constant ,we always use molarity as factors. So,

$\;\sf{\Longrightarrow\;\;Equilibrium\;Constant\;=\;\bf{\dfrac{[C]\:[D]}{[A]\:[B]}}}$

By applying values, we get

$\;\sf{\Longrightarrow\;\;Equilibrium\;Constant\;=\;\bf{\dfrac{\bigg[\bigg(\dfrac{0.666}{V}\bigg)\bigg]\:\bigg[\bigg(\dfrac{0.666}{V}\bigg)\bigg]}{\bigg[\bigg(\dfrac{0.334}{V}\bigg)\bigg]\:\bigg[\bigg(\dfrac{0.334}{V}\bigg)\bigg]}}}$

$\;\sf{\Longrightarrow\;\;Equilibrium\;Constant\;=\;\bf{\dfrac{\bigg(\dfrac{0.666}{V}\bigg)\:\bigg(\dfrac{0.666}{V}\bigg)}{\bigg(\dfrac{0.334}{V}\bigg)\:\bigg(\dfrac{0.334}{V}\bigg)}}}$

$\;\sf{\Longrightarrow\;\;Equilibrium\;Constant\;=\;\bf{\dfrac{\bigg(\dfrac{0.666}{V}\bigg)^{2}}{\bigg(\dfrac{0.334}{V}\bigg)^{2}}}}$

$\;\sf{\Longrightarrow\;\;Equilibrium\;Constant\;=\;\bf{\dfrac{\dfrac{(0.666)^{2}}{V^{2}}}{\dfrac{(0.334)^{2}}{V^{2}}}}}$

Cancelling V², we get

$\;\sf{\Longrightarrow\;\;Equilibrium\;Constant\;=\;\bf{\dfrac{(0.666)^{2}}{(0.334)^{2}}}}$

$\;\sf{\Longrightarrow\;\;Equilibrium\;Constant\;=\;\bf{\dfrac{0.443556}{111556}}}$

$\;\sf{\Longrightarrow\;\;Equilibrium\;Constant\;=\;\bf{\red{3.97\;\approx\;4}}}$

$\;\sf{\Longrightarrow\;\;Equilibrium\;Constant\;=\;\bf{\green{4}}}$

This is the required answer.

So the correct option is d.) 4

$\;\underline{\boxed{\tt{Required\;\: equilibrium\;\;Constant\;=\;\bf{\purple{4}}}}}$

________________________________

★ More to know :-

• Formulas related to Thermodynamics :-

$\;\tt{\leadsto\;\;\Delta U\;=\;q\;+\;W}$

$\;\tt{\leadsto\;\;\Delta U\;=\;\dfrac{q}{T}}$

$\;\tt{\leadsto\;\;\Delta_{r}G\;=\;\Delta_{r}H\;-\;T\Delta_{r}S}$

Answer 2

Explanation:

★ Concept :-

Here the concept of Equilibrium has been used. We see that we are given the initial moles of ethyl alcohol and acetic acid. So firstly from the number of moles of ester at equilibrium, we can find the number of moles of water, acetic acid and ethyl alcohol. And then by using the formula of Equilibrium Constant, we can find the required value.

Let's do it !!

______________________________________

★ Formula Used :-

For a reaction,

A + B ⇌ C + D

$\;\boxed{\sf{\pink{Equilibrium\;Constant\;=\;\bf{\dfrac{[C]\:[D]}{[A]\:[B]}}}}}$

______________________________________

★ Solution :-

Given reaction ,

→ C₂H₅OH + CH₃COOH ⇌ CH₃COOC₂H₅ + H₂O

Here ,

» Initial moles of C₂H₅OH = 1 mole

» Initial moles of CH₃COOH = 1 mole

Since when the reaction is at equilibrium,

At equilibrium, moles of C₂H₅OH = n

Similarly, since both of them are equal initially, so at equilibrium, moles of CH₃COOH = n

At equilibrium, moles of CH₃COOC₂H₅ = 0.666 moles = e (say)

Since water is also formed which is also at equilibrium, so at equilibrium, moles of H₂O = 0.666 moles = e (say)

-----------------------------------------------------------

~ For the value of n ::

Since those all reactants and products are in equilibrium. So at equilibrium their number of moles will be given as,

⟩⟩ n = 1 - e

⟩⟩ n = 1 - 0.666

⟩⟩ n = 0.334

Hence,

At equilibrium, number of moles of C₂H₅OH = 0.334

At equilibrium, number of moles of CH₃COOH = 0.334

-----------------------------------------------------------

~ For the concentration of reactants ::

Since we see that the number of moles of reactants are equal and number of moles of products are also equal. So their molarity (concentration) per volume will be also equal. So,

$\;\bf{\rightarrow\;\;\orange{Molarity\;=\;\dfrac{Number\;of\;Moles}{Volume\;of\;Solution}}}$

• For reactants ::

By applying values in formula, we get

$\;\bf{\rightarrow\;\;Molarity\;=\;\dfrac{0.334}{V}}$

• For Products ::

By applying values, we get

$\;\bf{\rightarrow\;\;Molarity\;=\;\dfrac{0.666}{V}}</p><p>$

-----------------------------------------------------------

~ For Equilibrium Constant ::

In finding equilibrium Constant ,we always use molarity as factors. So,

$\;\sf{\Longrightarrow\;\;Equilibrium\;Constant\;=\;\bf{\dfrac{[C]\:[D]}{[A]\:[B]}}$

By applying values, we get

$\;\sf{\Longrightarrow\;\;Equilibrium\;Constant\;=\;\bf{\dfrac{\bigg[\bigg(\dfrac{0.666}{V}\bigg)\bigg]\:\bigg[\bigg(\dfrac{0.666}{V}\bigg)\bigg]}{\bigg[\bigg(\dfrac{0.334}{V}\bigg)\bigg]\:\bigg[\bigg(\dfrac{0.334}{V}\bigg)\bigg]}}}$

$\;\sf{\Longrightarrow\;\;Equilibrium\;Constant\;=\;\bf{\dfrac{\bigg(\dfrac{0.666}{V}\bigg)\:\bigg(\dfrac{0.666}{V}\bigg)}{\bigg(\dfrac{0.334}{V}\bigg)\:\bigg(\dfrac{0.334}{V}\bigg)}}}$

$\;\sf{\Longrightarrow\;\;Equilibrium\;Constant\;=\;\bf{\dfrac{\bigg(\dfrac{0.666}{V}\bigg)^{2}}{\bigg(\dfrac{0.334}{V}\bigg)^{2}}}}$

$\;\sf{\Longrightarrow\;\;Equilibrium\;Constant\;=\;\bf{\dfrac{\dfrac{(0.666)^{2}}{V^{2}}}{\dfrac{(0.334)^{2}}{V^{2}}}}}$

$\;\sf{\Longrightarrow\;\;Equilibrium\;Constant\;=\;\bf{\dfrac{(0.666)^{2}}{(0.334)^{2}}}}$

$\;\sf{\Longrightarrow\;\;Equilibrium\;Constant\;=\;\bf{\dfrac{0.443556}{111556}}$

$\;\sf{\Longrightarrow\;\;Equilibrium\;Constant\;=\;\bf{\red{3.97\;\approx\;4}}}$

$\;\sf{\Longrightarrow$

;\;Equilibrium\;Constant\;=\;\bf{\green{4}}[/tex]

}⟹EquilibriumConstant=4

This is the required answer.

So the correct option is d.) 4

$\;\underline{\boxed{\tt{Required\;\: equilibrium\;\;Constant\;=\;\bf{\purple{4}}}}}$

________________________________

★ More to know :-

• Formulas related to Thermodynamics :-

$</p><p>\;\tt{\leadsto\;\;\Delta U\;=\;q\;+\;W}⇝ΔU=q+W$

$</p><p>\;\tt{\leadsto\;\;\Delta U\;=\;\dfrac{q}{T}}⇝ΔU=$

$</p><p>\;\tt{\leadsto\;\;\Delta_{r}G\;=\;\Delta_{r}H\;-\;T\Delta_{r}S}⇝Δ$