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 chemical equilibrium relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases

The equilibrium constant $K p$ for the reaction $2H _2(g) + O _2(g)  \rightleftharpoons 2H _2O(g)$ at 2000 K is $1.6 \times 10^7$. The equilibrium constant of the reaction $H _2O(g)  \rightleftharpoons H _2(g) + \frac{1}{2} O _2(g)$ is _______.

  1. $6.25 \times 10^{-8}$
  2. $2.5 \times 10^{-4}$
  3. $7.5 \times 10^{-12}$
  4. $4\times 10^{-6}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the reaction, $2H _2(g) + O _2(g)  \rightleftharpoons 2H _2O(g)$,
$K = \dfrac{{[H _2O]}^{2} _{}} {{[H _2]}^{2} _{} [{O} _{2}]}$

$2H _2O(g)  \rightleftharpoons 2H _2(g) + O _2(g)$
$K = \dfrac{{[H _2]}^{2} _{} [{O} _{2}]}{[{H _2O}]^{2} _{}}$

$H _2O(g)  \rightleftharpoons H _2(g) + \frac{1}{2} O _2(g)$
$K _1 = \sqrt{\dfrac{{[H _2O]}^{2} _{}} {{[H _2]}^{2} _{} [{O} _{2}]}}$

Thus, $K _1 = \dfrac{1}{\sqrt{1.6107}}$$= 2.5 \times 10^{-4}$

Multiple choice chemistry chemical equilibrium relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases

$K {c}$ for an equilibrium $SO _{3}\rightleftharpoons SO _{2}(g)+ \frac{1}{2}O _{2}(g)$ is equal to 0.15 at 900 K. The equilibrium constant for the equation $2SO _{2}+ O _{2}\rightleftharpoons 2SO _{3}(g)$ is ___________.

  1. $6.66$
  2. $44.4$
  3. $2.25$
  4. $0.44$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$K _{c}$ for $SO _{3}\rightleftharpoons  SO _{2}+ \frac{1}{2} O _{2}$ is 0.15.

$K _{c}$ for the following reaction $ 2SO _{2}+ O _{2}\rightleftharpoons 2SO _{3}$ will be the reciprocal of square of the above equilibrium as stoichiometric coefficients are doubled and the reaction is reversed.

$K^{1} _{c} = \dfrac{1}{K^{2} _{c}} = \dfrac{1}{0.15^{2}} = 44.4$

Multiple choice chemistry chemical equilibrium relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases

Calculate value of $'ln(K _{eq})$' for the reaction at 250 K.
$N _2O _4 (g) \rightleftharpoons 2NO _2 (g)$
Given: $H^0 _f((NO _2)g) = + 40.407 kJ / mol$
$H^0 _f((N _2O _4)g) = + 70 kJ / mol$
$S^0 _r = 10 JK^{-1}$

  1. 4

  2. -4

  3. 1.2

  4. -1.2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$+G^0 = +H^0 -TS^0$
$+H _r^0 = (+ H _f^0)p - ( + H _f^0) _R$
$= 2 \times (+ 40.407) - (+70)$
$= + 10.814 kJ$
$+G^0 = + 10.814 kJ - 250 \times 10 J/K$
$= + 10.814 kJ - 2500 J = 8314 J$
$+G^0 = -RT ln K$
$8314 = - 8.314 \times 250 ln K$
$ln K = - 4.$

Multiple choice chemistry chemical equilibrium relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases

The value of equilibrium constant $(K _f)$ for the reaction: $Zn^{2+}(aq)+4OH^{-}(aq)\rightleftharpoons Zn(OH) _{4}^{2-}(aq)$ is represented in scientific notation as $p\times10^{q},$ then q is:
Given : $Zn^{2+}(aq)+2e^{-}\rightarrow Zn(s);  E^{0}=-0.76 V$
            $Zn(OH) _{4}^{2-}(aq)+2e^{-}\rightarrow Zn(s)+4OH^{-}(aq);  E^{0}=-1.36V$
            $2.303\dfrac{RT}{F}=0.06$

  1. 20

  2. 10

  3. 15

  4. 21
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\Delta G^o=-RTlnK _{eq} $

$logK _{eq}=\dfrac{nFE^{o}}{RT\times 2.303}\Rightarrow \dfrac{2\times 0.6}{0.06}\Rightarrow 20$

$K=10^{20}$

Multiple choice chemistry chemical equilibrium relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases


Greenhouse gas $CO _2$ can be converted to $CO(g)$ by the following reaction

$CO _2(g)+H _2(g)\rightarrow CO(g)+H _2O(g)$,
 
termed as water gas reaction.

The Equilibrium constant $K _p$ for the water gas reaction at $1000\;K$ is: 

 $(\Delta H _{\displaystyle1000\;K}=35040\;J\;mol^{-1}\ ; \Delta S _{1000\;K}=32.11\;J\;mol^{-1}\ K^{-1})$

(Note : The gases behave ideally).

  1. $K _p=0.7030$
  2. $K _p=0.7300$
  3. $K _p=0.7330$
  4. $K _p=0.7303$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the formula Delta G = Delta H - T*Delta S, we find Delta G = 35040 - (1000 * 32.11) = 35040 - 32110 = 2930 J/mol. Then, using Delta G = -RT ln(Kp), ln(Kp) = -2930 / (8.314 * 1000) = -0.3524. Kp = exp(-0.3524) approximately 0.703.

Multiple choice chemistry chemical equilibrium relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases

Steam undergoes decomposition at high temperature as per the reaction
$H _{2}O(g) \rightleftharpoons  H _{2}(g)+\frac{1}{2}O _{2}(g), \Delta H^{\circ}=200 kJ  mol^{-1} \Delta S^{\circ}=40  J  mol^{-1}$
The temperature at which equilibrium constant is unit is :

  1. 3000 Kelvin

  2. 5000 Kelvin

  3. 5333 Kelvin

  4. 5 Kelvin
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}=-RT  ln Keq = 0  [Keq = 1]$.
$\Rightarrow T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{200\times 10^{3}}{40}=5000  K$

Multiple choice chemistry chemical equilibrium relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases

1 mole of $PCl-{3}$ and 1 mole of $PCl _{5}$ is taken in a vessel of 10 L capacity maintained at 400 K.At equilibrium, the moles of $Cl _{2}$ is found to be $4\times10^{-3}$

  1. $K _{c}$ for the reaction :$PCl _{5}(g)$$\rightleftharpoons$$PCl _{3}(g)+Cl _{2}(g) is 4\times10^{-4}$ M.
  2. $K _{p}$ for the reaction :$PCl _{3}+Cl _{2}(g)$$\rightleftharpoons$$PCl _{5}(g)+(g) is 4\times10^{-4}\times(0.082\times400)$ atm
  3. If $PCl _{3}(g)$ is added to the equilibrium mixture,$K _{p}$ at the new equilibrium becomes greater than the $K _{p}$ at old equilibrium.
  4. After equilibrium is achieved , moles of $PCl _{3}$ is doubled and moles of $Cl _{2}$ is halved simultaneously then the partial pressure of $PCl _{5}$remain unchanged.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the reaction PCl5 <=> PCl3 + Cl2, Kc = [PCl3][Cl2] / [PCl5]. Initial moles: PCl3=1, PCl5=1. At equilibrium, [Cl2] = 4*10^-3 / 10 = 4*10^-4 M. [PCl3] = (1+4*10^-3)/10 = 0.1004 M. [PCl5] = (1-4*10^-3)/10 = 0.0996 M. Kc = (0.1004 * 4*10^-4) / 0.0996 approx 4*10^-4.

Multiple choice chemistry chemical equilibrium relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases

${ K } _{ P }$ for the reaction ${ N } _{ 2 }+{ 3H } _{ 2 }\rightleftharpoons { 2NH } _{ 3 }$ at 400C is $1.64\times { 10 }^{ -4 }$.  Find ${ K } _{ C }$. Also find ${ \triangle G }^{ \oplus  }$ using ${ K } _{ P }$ and ${ K } _{ C }$ values  and interpret the differences.

  1. ${ K } _{ C }= 0.025$
    ${ \triangle G }^{ \oplus }= + 11.733 kcal$
  2. ${ K } _{ C }= 2.001$
    ${ \triangle G }^{ \oplus }= + 19.249 kcal$
  3. ${ K } _{ C }= 0.5006$
    ${ \triangle G }^{ \oplus }= + 11.733 kcal$
  4. ${ K } _{ C }= 1.5$
    ${ \triangle G }^{ \oplus }= + 15.22 kcal$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\displaystyle  \Delta n = 2 - [1+3] = -2$
$\displaystyle  K _p = K _c (RT)^{\Delta n}$
$\displaystyle 1.64 \times 10^{-4} = K _c (0.08206 \times 673)^{-2} $
$\displaystyle K _c = 0.5006 $
$\displaystyle \Delta G^0 = -RTln K _p = - 2 \times 673 \times ln 1.64 \times 10^{-4} =  11731 cal/mol = 11.733 kcal/mol$
$\displaystyle  \Delta G^0 = -RTln K _c = -2 \times 673 \times ln 0.5006 = 932 cal/mol = 0.93 kcal/mol$
The standard free energy change calculated form $K _p$ is higher than the standard free energy change calculated from $K _c$ as the numerical value of $K _c$ is higher than the numerical value of $K _p$
Multiple choice chemistry chemical equilibrium relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases

One mole of a compound AB reacts with one mole of a compound CD according to the equation ${ AB } _{ \left( g \right) + }{ CD } _{ \left( g \right)  }\rightleftharpoons { AD } _{ \left( g \right)  }+{ CB } _{ \left( g \right)  }$ When equilibrium had been established it is was found that 3/4 mole of reactants AB and CD had been converted to AD and CB, there is no change in volume. The equilibrium constant for the reaction is :

  1. 9/16

  2. 1/9

  3. 16/9

  4. 9
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that


initial moles of $AB=1$ mol

initial moles of $CD=1$ mol

moles of $AB$ reacted $=\dfrac{3}{4}$ moles

moles of $AB$ left $=1-\dfrac{3}{4}=\dfrac{1}{4}$ mol

Similarly 

moles of $CD$ left $=1-\dfrac{3}{4}=\dfrac{1}{4}$ mol

Now,

$t=0\,\,1\,mol\,\,\,\,\,\,\,1\,mol\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0mol\,\,\,\,\,\,0mol$
          $AB(g)+CD(g)\,\,\,\,\rightleftharpoons  AD(g)+CB(g)$
teq     $\dfrac{1}{4}mol\,\,\,\,\,\,\dfrac{1}{4}mol\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{3}{4}mol\,\,\,\,\,\,\dfrac{3}{4}mol$

Now, 
$[AB]=\dfrac{n _{AB}}{v}=\dfrac{1/4}{v}M$  (where, v=volume)

$[CD]=\dfrac{n _{CD}}v{}=\dfrac{1/4}{v}M$

$[AD]=[CB]=\dfrac{3/4}{v}M$

Now $K _c=\dfrac{[AD][CB]}{[AB][CD]}=\dfrac{\dfrac{3/4}{v}M\times \dfrac{3/4}{v}M}{\dfrac{1/4}{v}M\times \dfrac{1/4}{v}M}$

$=3\times 3=9$

Multiple choice chemistry chemical equilibrium relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases

Hydrolysis of sucrose gives glucose and fructose. The reaction takes place as: Sucrose $+ H _{2}O \rightleftharpoons$ Glucose $+$ Fructose. The equilibrium constant $K _{c}$ for this reaction is $2\times 10^{13}$ at $300\ K$. The $ \Delta G^{\circ}$ at $300\ K$ is:

  1. $7.64\times 10^{4}\ J\ mol^{-1}$
  2. $7.64\times 10^{-4}\ J\ mol^{-1}$
  3. $-7.64\times 10^{-4}\ J\ mol^{-1}$
  4. $-7.64\times 10^{4}\ J\ mol^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The relationship between the standard free energy change $(\Delta G^o)$ and the equilibrium constant $(K _p)$ is $(\Delta G^o)=-RTlnK _c$, where, $R$ is the ideal gas constant and $T$ is the temperature.


Given, $K _c=2\times 10^{13}, T=300K, R=8.314 :Jmol^{-1}K^{-1}$

Substituting these values in the above expression, we get

$(\Delta G^o)=-RTlnK _c=-8.314 \times 300 \times ln(2\times 10^{13})=-7.64\times 10^{4} J mol^{-1}$ 

Hence, the standard free energy change $(\Delta G^o)=-7.64\times 10^{4} J mol^{-1}$

Multiple choice chemistry chemical equilibrium relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases

 $SO _2(g) + 1/2O _2 (g)\rightleftharpoons SO _3(g) \Delta H^o _{298} = 98.32 kJ/mole, \Delta S^o _{298} = 95.0 J/K/mole$.


 Find the $K _p$ for this above reaction at 298K:

  1. $K _P = 9.31 \times 10^{-12} atm^{1/2}$
  2. $K _P = 5.34 \times 10^{-13} atm^{1/2}$
  3. $K _P = 3.7 \times 10^{-13} atm^{1/2}$
  4. $K _P = 3.7 \times 10^{-14} atm^{1/2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\displaystyle \Delta G^0 = \Delta H^0 - T\Delta S^0  $
$\displaystyle  \Delta G^0 =  98.32 \times 1000 - 298 \times 95.0 = 70010 J/mol$
$\displaystyle  \Delta G^0 = -RTlnK _P$
$\displaystyle 70010 = - 8.314 \times 298 \times ln K $
$\displaystyle  ln K = -28.26$
$\displaystyle  K = 5.34 \times 10^{-13}$
Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

A sample of $CaC{O _3}$ is $50\% $ pure. On heating $1.12{\text{ }}L$ of $C{O _2}$ (at STP) is obtained. Residue left (assuming non-volatile impurity) is:

  1. 7.8 g

  2. 3.8 g

  3. 2.8 g

  4. 8.9 g
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

No. of moles of $CO _2$ evolved $=\cfrac{1.12}{22.4}=0.05$ $moles$

$CaCO _3(s)\overset { \Delta }{ \longrightarrow } CaO\downarrow+CO _2\uparrow$
                             $0.05$        $0.05$ $moles$
So, $0.05$ $moles$ of $CaCO _3$ have  reacted.
Mass $=0.05\times 100=5$ $gm=50\%$ of $CaCO _3$ sample
Total weight $=2\times 5$ $gm=10$ $gm$
Residue left by $CaCO _3=5$ $gm$
Residue left by $CaO=56\times 0.05=2.8$ $gm$
Toatl residue $=5+2.8=7.8$ $gm$

Multiple choice chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

In the reaction ${ N } _{ 2 }+{ 3H } _{ 2 }\longrightarrow { 2H } _{ 3 }$, ratio by volume of ${ N } _{ 2 },{ H } _{ 2 }$ and $ { NH } _{ 3 }$ is $1:3:2$. This illustrates law of :

  1. definite proportions

  2. multiple proportions

  3. reciprocal proportions

  4. gaseous proportions
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As per the Gay Lussac's law of combining volume of gases, the volumes of gaseous reactants and gaseous products bear a simple whole number ratio with each other if they are measured at same temperature and pressure.
In the reaction ${ N } _{ 2 }+{ 3H } _{ 2 }\longrightarrow { 2H } _{ 3 }$, ratio by volume of $N _2,  H _2$ and $ NH _3$ is $1:3:2$. This illustrates law of Gay Lussac's law of combining volumes of gases.

Multiple choice chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

In the reaction $N _{2}+3H _{2}\rightarrow 2NH _{3} $, the ratio by volume of $N _{2},\ H _{2} :$ and$: NH _{3}$ is $1 : 3 : 2$. 


This illustrates the law of:

  1. definite proportion

  2. multiple proportion

  3. reciprocal proportion

  4. gaseous volumes
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In the reaction, $N _{2}+3H _{2}\rightarrow 2NH _{3} $, the ratio by volume of $N _{2},\ H _{2} $ and $ NH _{3}$ is $1 : 3 : 2$. This illustrates the law of Gaseous volumes or Gay Lussac's law of combining volumes of gases.

According to this law, when gases react together to produce gaseous products, the volume of reactants and products bear a simple whole-number ratio with each other, provided volumes are measured at the same temperature and pressure.

So, the correct option is $D$.