Chemistry

Chemical Equilibrium and Stoichiometry

103 Questions

Chemical equilibrium and stoichiometry questions cover equilibrium constants, percent yield, and the Haber process. These concepts test your ability to calculate reaction outcomes and purity. Practice these to excel in physical chemistry sections of competitive tests.

Equilibrium constant calculationsStoichiometric yieldHaber processPercent purityHomogenous equilibrium

Chemical Equilibrium and Stoichiometry Questions

Multiple choice chemistry further aspects of equilibria indicators and acid-base titration study of indicators properties of acids and bases

In the mixture of $NaHCO _{4}$ and $Na _{2}CO _{3}$, volume of a given $HCl$ required is $x\ mL$ with phenolphthalein indicator and $y\ mL$ with methyl orange indicator in same titration. Hence, volume of $HCl$ for complete reaction of $Na _{2}CO _{3}$ present in the original mixture is

  1. $2x$
  2. $y$
  3. $x/2$
  4. $(y - x)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

With phenolphthalein, Na2CO3 is converted to NaHCO3 (half neutralization). With methyl orange, the NaHCO3 is converted to NaCl. The volume for the second half is equal to the first, so the total volume for Na2CO3 is 2x.

Multiple choice chemistry further aspects of equilibria indicators and acid-base titration study of indicators properties of acids and bases

$40\ mL$ of $0.05\ M\ Na {2}CO _{3}\cdot NaHCO _{3} \cdot 2H _{2}O$ (sesquicarbonate) is titrated against $0.05\ M\ HCl.\ x\ mL$ of $HCl$ is used when phenolphthalein is the indicator and $y\ mL\ HCl$ is used when methyl orange is the indicator in two separate titrations, hence $(y - x)$ is_______.

  1. $80\ mL$
  2. $30\ mL$
  3. $120\ mL$
  4. none of the above

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Titration of ${ Na } _{ 2 }{ CO } _{ 3 }.{ NaHCO } _{ 3 }.2{ H } _{ 2 }O$ with $HCl$ involves following reactions :
a) ${ Na } _{ 2 }{ CO } _{ 3 }+HCl\rightleftharpoons { NaHCO } _{ 3 }+NaCl$
b) ${ NaHCO } _{ 3 }+HCl\rightleftharpoons NaCl+{ H } _{ 2 }O+{ CO } _{ 2 }$
In step $a$, $40$ ml of $0.05M$ $HCl$ will react with $40$ ml of $0.05M$ ${ Na } _{ 2 }{ CO } _{ 3 }$ to form ${ NaHCO } _{ 3 }$ using phenolpthalein.
$\therefore$   $x=40$ ml
Now, in a separate titration $40$ ml of $0.05M$ $HCl$ will need to react with $0.05M$ ${ Na } _{ 2 }{ CO } _{ 3 }$ to form ${ NaHCO } _{ 3 }$. Now in the second step $b$ total $80$ ml of $0.05M$ $HCl$ will need to neutralise ${ NaHCO } _{ 3 }$ completely.
$\therefore$   $y=40+40\times 2=120$
$\therefore$   $y-x=120-40=80$ ml
Answer will be $A$.
Multiple choice chemistry study of compounds a. hydrogen chloride laboratory method of preparation of hydrochloric acid physical and chemical properties and uses of hydrochloric acid hydrogen chloride: occurrence, preparation, purification and identification

In which reaction equilibrium moves in left hand side when pressure is increased?

  1. $H _{2(g)} + Cl _{2(g)} \rightleftharpoons 2HCl _{(g)}$
  2. $2Mg _{(s)} + O _{2(g)} \rightleftharpoons 2MgO _{(s)}$
  3. $2H _{2}O _{(g)} \rightleftharpoons 2H _{2(g)} + O _{2(g)}$
  4. $N _{2(g)} + 3H _{2(g)} \rightleftharpoons 2NH _{3(g)}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to Le Chatelier's principle, increasing pressure shifts the equilibrium toward the side with fewer moles of gas. In reaction C, there are 2 moles of gas on the left and 3 moles on the right. Increasing pressure shifts it to the left (fewer moles).

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

$4g \,H _2$ 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:

  1. $3.25$
  2. $1.75$
  3. $2.25$
  4. $2.5$
Reveal answer Fill a bubble to check yourself
D Correct answer
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,
$=\left(2-\cfrac {0.05}{2}\right)+\left(0.5-\cfrac {0.05}{2}\right)+0.05$
$=2.5$

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

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:

  1. Adding a suitable catalyst

  2. Adding an inert gas

  3. Decreasing the volume of container

  4. Increasing the amount of $CO(g)$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $CO _2$.

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

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_______.

  1. PCl$ _5$ will remain unchanged
  2. Cl$ _2$ will be greater
  3. PCl$ _5$ will become less
  4. PCl$ _5$ will become greater
Reveal answer Fill a bubble to check yourself
C Correct answer
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$.

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

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:

  1. $[AB _3] _{eq} = [BC] _{eq}$
  2. $[AB _2] _{eq} = [B _2C] _{eq}$
  3. $[AB _3] _{eq} > [B _3C _2] _{eq}$
  4. $[AB _3] _{eq} > [BC] _{eq}$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
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  $[AB _3] _{eq}  >  [BC] _{eq}$

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

The equilibrium constant for a reaction is $K$, and the reaction quotient is $Q$. For a reaction mixture, the ratio $\dfrac {K}{Q}$ is $0.33$. This means that:

  1. the reaction mixture will equilibrium to form more reactant species

  2. the reaction mixture will equilibrium to form more product species

  3. the equilibrium ratio of reactant to product concentrations will be $3$
  4. the equilibrium ratio of reactant to product concentrations will be $0.33$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that for a reaction if $\dfrac{K}{Q} < 1,$ the reaction proceeds in backward direction. Hence, the reaction mixture forms more reactant species.

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

Assume that the decomposition of $H{ NO } _{ 3 }$ can be represented by the following equation
$4H{ NO } _{ 3 }(g)\rightleftharpoons 4{ NO } _{ 2 }(g)+2{ H } _{ 2 }O(g)+{ O } _{ 2 }(g)\quad $'and the reaction approaches equilibrium at $400K$ temperature and $30$ atm pressure. The equilibrium partial pressure of $H{ NO } _{ 3 }$ is $2$ atm
Calculate ${K} _{c}$ in ${ \left( mol/L \right)  }^{ 3 }$
(Use: $R=0.08atm-L/mol-K$)

  1. $4$
  2. $8$
  3. $16$
  4. $32$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the relation between Kp and Kc, namely Kp = Kc * (RT)^delta_n, we find delta_n = (4 + 2 + 1) - 4 = 3. Given P(HNO3) = 2 atm and total pressure 30 atm, we can find partial pressures at equilibrium, then compute Kp and subsequently Kc.

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

The optical rotation of the $\alpha-form$ of a pyramose is $+150.7^{\circ}$, that of the $\beta - form$ is $+52.8^{\circ}$. In solution an equilibrium mixture of these anomers has an optical rotation of $+80.2^{\circ}$. The percentage of the $\alpha$ form in equilibrium mixture is:

  1. $28$%
  2. $32$%
  3. $68$%
  4. $72$%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\alpha $ from $=+150.{ 7 }^{ 0 }$, $\beta $ from $=+52.{ 8 }^{ 0 }$

at equilibrium optical rotation $=+80.2$ 
let, at equilibrium $\alpha $-from exist $=x$
      at equilibrium $\beta $-from exist $=(100-x)$
Therefore, $\dfrac { 150.7x+\left( 100-x \right) \times 52.8 }{ 100 } =80.2$
$\Rightarrow \quad 150.7x+5280-52.8x=8020$
$\Rightarrow \quad 99.9x=2740$
$\Rightarrow \quad x=27.42\approx 28$%

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

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 $(Q _{c})$?

  1. $\dfrac {3}{2}$
  2. $\dfrac {2}{3}$
  3. $1$
  4. None of these

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

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.

  1. $1.8 \times 10^2$
  2. $2.8 \times 10^2$
  3. 0.034

  4. $2.8 \times 10^{-2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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:

  1. $2.5 \times 10^{-3}$
  2. $2.5 \times 10^{3}$
  3. $1.0 \times 10^{-5}$
  4. $5 \times 10^{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To obtain the target equilibrium equation, multiply the first reaction by -1/2 (or reverse and take square root) and add the second reaction. This corresponds to combining the equilibrium constants as K3 = (1 / sqrt(K1)) * K2.

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

Which of the following statements is correct?

  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.
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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.

  1. $2.96$
  2. $6.14$
  3. $204.8$
  4. None of these

Reveal answer Fill a bubble to check yourself
A Correct answer
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$^3$.