Tag: introduction to equilibrium

Questions Related to introduction to equilibrium

A reaction mixture containing $H _{2}, N _{2}$ and $NH _{3}$ has partial pressure $2\ atm, 1\ atm$ and $3\ atm$ respectively at $725\ K$. If the value of $K _{P}$ for reaction, $N _{2} + 3H _{2}\rightleftharpoons 2NH _{3}$ is $4.28\times 10^{-5} atm^{-2}$ at $725\ K$, in which direction the net reaction will go :

chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

  1. Forward

  2. Backward

  3. No net reaction

  4. Direction of reaction cannot be predicted

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Correct Option: A
Explanation:

                ${ N } _{ 2 }+{ 3H } _{ 2 }\rightleftharpoons 2{ NH } _{ 3 }$

at eqn,   1atm    2atm       3atm
given, ${ K } _{ p }=4.28\times { 10 }^{ -5 }{ atm }^{ -2 }$
${ Q } _{ p }=\dfrac { { \left( { P } _{ { NH } _{ 3 } } \right)  }^{ 2 } }{ \left( { P } _{ { N } _{ 2 } } \right) { \left( { P } _{ { H } _{ 2 } } \right)  }^{ 3 } } =\dfrac { { \left( 3 \right)  }^{ 2 } }{ 1\times { \left( 2 \right)  }^{ 3 } } =\dfrac { 3\times 3 }{ 2\times 2\times 2 } =\dfrac { 9 }{ 8 } =1.125{ atm }^{ -2 }$
Since, $\boxed { { Q } _{ p }>>{ K } _{ p } } $
Reaction will move in forward direction.

Which of the following is true :

chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

  1. $pk _{b}$ for $OH^{-}$ is -1.74 at $25^{o}$C

  2. The equilibrium constant for the reaction between HA ($pk _{a} = 4$) and NaOH at $25^{o}$C will be equal to $10^{10}$

  3. The pH of a solution containing 0.1 M HCOOH ($k _{a} = 1.8 \times 10^{-4}$) and 0.1 M HOCN ($k _{a} = 3.2 \times 10^{-4}$) will be nearly (3 -log 7)

  4. All of the above are correct

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Correct Option: A

A mixture of three gases P (density 0.90), Q (density 0.178) and R (density 0.42) is enclosed in a vessel at the constant temperature. When the equilibrium is established:

chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

  1. the gas P will be at the top of the vessel

  2. the gas Q will be at the top of the vessel

  3. the gas R will be at the top of the vessel

  4. the gases will mix homogeneously throughout the vessel.

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Correct Option: D
Explanation:

Independent of density of gases, in equilibrium gases will mix homogenously, this comes from the fact that gases occupies entire volume of container. This also can be thought as gases tries to reduce energy. Hence they separate as far as possible.

The equilibrium constant $K _{c}$ for the reaction $P _{4}(g) \rightleftharpoons 2P _{2}(g)$
is $1.4$ at $400^{\circ}C$. Suppose that $3$ moles of $P _{4}(g)$ and $2$ moles of $P _{2}(g)$ are mixed in $2$ litre container at $400^{\circ}C$. What is the value of reaction quotient $(Q _{c})$?

chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

  1. $\dfrac {3}{2}$

  2. $\dfrac {2}{3}$

  3. $1$

  4. None of these

Show answer
Correct Option: B
Explanation:

$Q _c = \dfrac{[P _2(g)]^4}{[P _4 (g)]}$

= $\dfrac{(1)^2}{(3/2)}$ 
= $\dfrac{2}{3}$

0.1 mole of $N _2O _4(g)$ was sealed in a tube under one atmospheric conditions at $25^0C$. Calculate the number of moles of $NO _2(g)$ present, if the equilibrium $N _2O _4(g)\rightleftharpoons  2NO _2(g)$ $(K _p=0.14)$ is reached after some time.

chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

  1. $1.8 \times 10^2$

  2. $2.8 \times 10^2$

  3. 0.034

  4. $2.8 \times 10^{-2}$

Show answer
Correct Option: C
Explanation:

According to the equation

$\begin{matrix}  & N _2O _4& \rightleftharpoons & 2NO _2 \ \text{Initial}&1 \, atm  & &0 \ \text{change}& -x&  &+2x\ \text{Equilibrium}&1-x&&2x  \end{matrix}$
$K _P = \dfrac{(P _{NO _2})^2}{P _{N _2O _4}}$   $[\therefore K _P$ is similar to $K _C$ in aspect of setting up rate quotient]

$0.14 = \dfrac{(2x)^2}{1-x}$

$0.14(1-x) = 4x^2$
$x = 0.17$
From ideal gas equation
$PV = nRT$

$V = \dfrac{nRT}{P}$

$V = \dfrac{0.1\times 0.082 \times (273 + 25)}{1}$

$V = 2.45 lit$

Moiles of $NO _2= \dfrac{P _{NO _2} \times V}{RT}$

Moles of $NO _2 = \dfrac{0.34\times 2.45}{0.0821 \times (273 + 25)}$

Moles of $NO _2 = 0.034$

$\therefore$ option C is correct

Consider the following reactions in which all the reactants and the products are in 


$2PQ \rightleftharpoons P _2 + Q _2 ; K _1 = 2.5 \times 10^5$

$PQ + \cfrac{1}{2} R _2 \rightleftharpoons PQR; K _2 = 5 \times 10^{-3}$

The value of $K _3$ for the equilibrium $\cfrac{1}{2} P _2 + \cfrac{1}{2} Q _2 + \cfrac{1}{2} R _2 \rightleftharpoons PQR,$ is:

chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

  1. $2.5 \times 10^{-3}$

  2. $2.5 \times 10^{3}$

  3. $1.0 \times 10^{-5}$

  4. $5 \times 10^{3}$

Show answer
Correct Option: A

Chemical equilibria is (I) _________ in nature. It occurs in (2) ________ reactions only.  At equilibria, all (3) __________ properties becomes constant. Chemical equilibrium gets affected by change in temperature, pressure, concentration, volume etc but is not altered by addition of (4)________. Equilibria are of two types (5) ________ and homogeneous equilibria.

chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

  1. (1) dynamic (2) reversible (3) observable (4) catalyst (5) homogeneous 

  2. (1) static (2) irreversible (3) observable (4) reactants (5) heterogeneous

  3. (1) dynamic (2) irreversible (3) observable (4) reactants (5) homogeneous 

  4. (1) dynamic (2) reversible (3) observable (4) catalyst (5) heterogeneous 

Show answer
Correct Option: D
Explanation:

Chemical equilibria is dynamic in nature. It occurs in reversible reactions only. At equilibria all observable properties becomes constant. Chemical equilibrium is affected by a change in temperature, pressure, concentration, volume etc but is not altered by the addition of catalyst. Equilibria is of two types heterogeneous and homogeneous equilibria.

Which of the following statements is correct?

chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

  1. In equilibrium mixture of ice and water kept in perfectly insulated flask, mass of ice and water does not change with time.

  2. The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate.

  3. On addition of catalyst, the equilibrium constant value is not affected.

  4. Equilibrium constant for a reaction with negative $\triangle$H value decreases as the temperature increases.

Show answer
Correct Option: B
Explanation:

The equilibrium reaction containing $Fe(III)$ nitrate & $KSCN$ is given as :-

${ Fe }^{ 3+ }\left( aq \right) +{ SCN }^{ - }\left( aq \right) \rightleftharpoons { \left[ Fe\left( SCN \right)  \right]  }^{ 2+ }\left( aq \right) $
Here, the red colour in the reactions occurs due to formation of ${ \left[ Fe\left( SCN \right)  \right]  }^{ 2+ }$. Now, when oxalic acid is added, it will react with ${ Fe }^{ 3+ }$ to form ${ \left[ Fe{ \left( { C } _{ 2 }{ O } _{ 4 } \right)  } _{ 3 } \right]  }^{ 3- }$. So, the concentration of ${ Fe }^{ 3+ }$ ions will get decreased. By applying Lechatelier's principle we see that the reaction will proceed towards backward direction and concentration of ${ \left[ Fe\left( SCN \right)  \right]  }^{ 2+ }$ will get decreased. Consequently the intensity of red colour is decreased.

When sulphur ( in the form of $S _8$) is heated to temperature T, at equilibrium, the pressure of $S _8$ falls by 30% from 1.0 atm, because $S _8$(g) is partially converted into $S _2$(g). Find the value of $K _p$ for this reaction.

chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

  1. $2.96$

  2. $6.14$

  3. $204.8$

  4. None of these

Show answer
Correct Option: A
Explanation:

                                                  ${{S} _{8}}\rightleftharpoons 4{{S} _{2}}$

Initial pressure                          $1$atm     $0$

At equilibrium pressure    $(1-0.3)$     $4\times 0.3$


Therefore, equilibrium pressure of ${{S} _{8}}=(1-0.3)=0.7$ atm

And equilibrium pressure of ${{S} _{2}}=4\times 0.3=1.2$ atm


So, equilibrium constant $Kp=\dfrac{P _{S _2}^4}{P _{S _8}}=\dfrac{{{1.2}^{4}}}{0.7}=$ approximate $2.96$ atm$^3$. 

In a dynamic equilibrium, the concentrations of reactants and products remains ___________.

chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

  1. differs with substance

  2. changes

  3. constant

  4. equilibrium

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Correct Option: C
Explanation:

In chemistry, a dynamic equilibrium exists once a reversible reaction ceases to change its ratio of reactants/products, but substances move between the chemicals at an equal rate, meaning there is no net change. It is a particular example of a system in a steady state. In thermodynamics a closed system is in thermodynamic equilibrium when reactions occur at such rates that the composition of the mixture does not change with time. Reactions do in fact occur, sometimes vigorously, but to such an extent that changes in composition cannot be observed.

Hence, in a dynamic equilibrium, the concentrations of reactants and products remains constant.