Tag: chemical equilibrium and acids-bases

Questions Related to chemical equilibrium and acids-bases

Gaseous $ N _{2}O _{4} $ dissociates into gaseous $ NO _{2} $ according to the reaction $ N _{2}O _{4} (g) \rightleftharpoons 2NO _{2}(g)$ at 300 K and 1 atm pressure, the degree of dissociation of $ N _{2}O _{4} $ is 0.2. If one mole of $ N _{2}O _{4} $ gas is contained in a vessel, then the density of the equilibrium mixture is : 

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

  1. 3.11 g/L

  2. 4.56 g/L

  3. 1.56 g/L

  4. 6.22 g/L

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Correct Option: A
Explanation:

$N _2O _4\longrightarrow 2NO _2$


at $t=0$, moles of $N _2O _4=1$, moles of $NO _2=0$

at $t=equilibrium$, mole of $N _2O _4=1-a$, mole of $NO _2=2a$

$a$=degree of dissociation.

Molecular weight of mixture$=\cfrac {(1-a)\times\text{molar mass of }N _2O _4+2a\times \text{molar mass of }NO _2}{(1-a+2a)}$
                                                $=\cfrac {(1-0.2)(28+64)+2\times 0.2\times (14+32)}{1+0.2}$
$M=76.66$

$P=1 atm,T=300K,$

$d=PM/RT$

    $=\cfrac {1\times 76.66}{0.082 \times 300}=3.11\  gm/lit\ $ .

Attainment of equilibrium in a coloured gaseous reversible reaction is detected by the constancy of:

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

  1. colour

  2. density

  3. pressure

  4. all the above properties of the mixture

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Correct Option: D
Explanation:

The intensity of color represents the concentration of either reactant or product. Thus when in a colored gaseous reversible reaction, the color has attained constant intensity, the concentrations of reactants and products have reached equilibrium values. In other words, an equilibrium is attained.


Hence, the correct option is A.

$4g \,H _2$ and $127g \,I _2$  are mixed and heated lit closed vessels until equilibrium is reached. If the equilibrium concentration of $HI$ is $0.05 \,M$ total number of moles present at equilibrium is:

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

  1. $3.25$

  2. $1.75$

  3. $2.25$

  4. $2.5$

Show answer
Correct Option: D
Explanation:

Given $H _2=4g$ & $I _2=127 g$

$\Rightarrow H _2=2$ mole & $I _2=0.5$ mole
$[HI] _{eqm}=0.05M$   $\therefore$ Moles of $HI=0.05$ mole
            $H _2\quad +\quad I _2\quad \rightleftharpoons\quad  2HI$
              $2$              $0.5$                  $0$         Initial
  $2-\cfrac {0.05}{2}$     $0.5-\cfrac {0.05}{2}$       $0.05$       Eqm
$\therefore$ Total moles at eqm,
$=\left(2-\cfrac {0.05}{2}\right)+\left(0.5-\cfrac {0.05}{2}\right)+0.05$
$=2.5$

For the reaction ${ CO(g)+H } _{ 2 }O(g)\rightleftharpoons { CO } _{ 2 }(g)+{ H } _{ 2 }(g)$ at a given temperature the equilibrium amount of ${ CO } _{ 2 }(g)$ can be increased by:

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

  1. Adding a suitable catalyst

  2. Adding an inert gas

  3. Decreasing the volume of container

  4. Increasing the amount of $CO(g)$

Show answer
Correct Option: D
Explanation:

$CO _(g)^+\ H _2(g)\rightleftharpoons CO _{2(g)}+H _{2(g)}$

$\Delta x=0$   $\therefore$ Adding inert gas & decreasing volume will have no effect. by increasing amount of CO, we shift reaction forward and to more $CO _2$.

Match the items in column - I with those in column - II

Column I Column II
1. Electric Fuse  A  Chemical Effect
2. Relay B Electric Discharge
3. CFL C Magnetic effect 
4. Button Cell D Heating Effect

Which of the following shows the correct matching ?

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

  1. 1 - C, 2 - B, 3 - A, 4 - D

  2. 1 - B, 2 - A, 3 - C, 4 - D

  3. 1 - D, 2 - C, 3 - B, 4 - A

  4. 1 - D, 2 - B, 3 - C, 4 - A

Show answer
Correct Option: C
Explanation:

Electric fuse is used to protect electric appliances. When current larger than specified value flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit.


 In relay transmission electromagnetic waves are used. 

CFL works on electric discharge. In button cell electrolytes are the sources of ions.

In the button cell electrolytes are sources of ions.

Option C is correct.

Equilibrium can be achieved only in open vessel.
chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

Equilibrium can be achieved in open vessel as well as in close vessel.

The reactions $PCl 5 (g)  \rightleftharpoons  PCl _3(g) + Cl _2 (g) $ and $COCl _2 (g)  \rightleftharpoons  CO(g) + Cl _2(g)$ are simultaneously in equilibrium in an equilibrium box at constant volume. A few moles of CO(g) are later introduced into the vessel. After some time, the new equilibrium concentration of_______.

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

  1. PCl$ _5$ will remain unchanged

  2. Cl$ _2$ will be greater

  3. PCl$ _5$ will become less

  4. PCl$ _5$ will become greater

Show answer
Correct Option: C
Explanation:

If CO is added 2$^{nd}$ equilibrium will proceed in the backward direction and concentration of Cl$ _2$ will decrease. This Cl$ _2$ will be further formed by the decomposition of PCl$ _5$.

If two gases $AB _2$ and $B _2C$ are mixed the following equilibria are readily established
$AB _2(g) + B _2 C(g)  \rightarrow AB _3(g) + BC(g)$
$BC(g) + B _2 C(g)  \rightarrow B _3 C _2 (g)$
If the reaction is started only with $AB _2$ with $B _2C$, then which of the following is necessarily true at equilibrium:

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

  1. $[AB _3] _{eq} = [BC] _{eq}$

  2. $[AB _2] _{eq} = [B _2C] _{eq}$

  3. $[AB _3] _{eq} > [B _3C _2] _{eq}$

  4. $[AB _3] _{eq} > [BC] _{eq}$

Show answer
Correct Option: C,D
Explanation:

Let reactions is started with a mole of $AB _2$ and b mole of $B _2C$
$\Rightarrow      AB _2 (g) + B _2C(g)  \rightarrow AB _3(g) + BC(g)$
                    a                 b               0               0
                  a - x           b - x - y        x              x - y
$BC(g) + B _2C(g)   \rightarrow B _2C _2 (g)$
    x - y            b - x - y        y                 As  x > y
Clearly $[AB _3] _{eq} > [B _3 C _2] _{eq} $ and  $[AB _3] _{eq}  >  [BC] _{eq}$

When the equilibrium is attained, the concentration of each of the reactants and products becomes equal.
chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

  1. True

  2. False

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Correct Option: B
Explanation:

When the equilibrium is attained, the concentration of each of the reactants and products may or may not become equal.  It may happen that the concentrations of products are higher than the concentrations of the reactants. Or, it may happen that the concentrations of the reactants are higher than the concentrations of the products.

At 373 K,a gaseous reaction $A\rightarrow 2B+C$ is found to be of first order.Starting with pure A,the total pressure at the end of 10 min was 176 mm of Hg and after a long time when A was completely dissociated,it was 270 mm of Hg.The pressure of A at the end of  10 minutes was:

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

  1. 94 mm of Hg

  2. 47 mm of Hg

  3. 43 mm of Hg

  4. 90 mm of Hg

Show answer
Correct Option: B
Explanation:

                    $A \to 2B + C$

$at\,t = 0$   $x$       $0$    $0$

$at\,t = 10$    $x-y$           $2y$    $y$  

 $total=x-y+2y+y$

        $=x+2y=176\,mm$-------$(i)$

$at\,{t={100}}$   $0$       $2x$    $x$

$total=2x+x=270\,mm$

$ \Rightarrow 3x = 270$
$ \Rightarrow x = 90\,mm\,\,of\,Hg$

put the value of $A$ in $e{q^n}\,(i),$ we get
   $x+2y=176$
$ \Rightarrow 90 + 2y = 176$
$ \Rightarrow  2y = 86$
$ \Rightarrow y=43\,\,\,mm\,\,of\,Hg$

At the end of $10$ min pessure of $A$ is  $x-y=90-43=47\,mm\,of\,Hg$

Option B is correct.