Chemistry

Chemical Equilibrium and Stoichiometry

135 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 equilibrium in physical processes introduction to equilibrium chemical equilibrium and acids-bases

Which of the following statements is incorrect ?

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

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

  3. On addition of catalyst 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
A Correct answer
Multiple choice chemistry chemical equilibrium equilibrium in physical processes introduction to equilibrium chemical equilibrium and acids-bases

Two flasks A and B of an equal volume containing 1 mole and 2 moles of O$ 3$ respectively are heated to the same temperature. When the reaction $2O _3 \rightleftharpoons 3O _2$ practically stops, then both the flasks shall have __________.

  1. the same ratio: $[O _2]/[O _3]$
  2. the same ratio: $[O _2]^{3/2}/[O _3]$
  3. only $O _2$

  4. the same time to reach equilibrium
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the given reaction,
$K _c = \displaystyle \frac{[O _2]^3}{[O _3]^2}= constant$     so  $\displaystyle \sqrt{K _c} = \frac{[O _2]^{3/2}}{[O _3]} = constant$
Same for both containers. Kc won't change as flasks are heated to same temperature. 

Multiple choice chemistry chemical kinetics introduction to chemical kinetics understanding chemical kinetics rate of chemical reaction

The approach to the following equilibrium was observed kinetically from both directions:
$ Pt{ Cl } _{ 4 }^{ -2 }+H _{ 2 }O\rightleftharpoons Pt(H _{ 2 }O)CI _{ 3 }+Cl^{ - } $
At $ 25^o C,$  it was found that  $ -\dfrac { d[PtC{ l } _{ 4 }^{ 2- }] }{ dt } =(3.9  \times 10^{ -5 } s^{ -1 }) [PtC{ l } _{ 4 }^{ 2- }]-(2.1  \times 10^{ -3 } L mol^{ -1 }s^{ -1 }) [Pt(H _{ 2 }O)C{ l } _{ 3 }^{ - }][Cl^{ - }] $
The value of $ K _{eq} $ (equilibrium constant) for the complexation of the fourth $ cl^- $ by $ Pt( II ) $ is

  1. $ 53.8 mol L^{-1} $
  2. $ 0.018 mol L^{-1} $
  3. $ 53.8 L mol^{-1} $
  4. $ 0.018 L mol^{-1} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

At equilibrium, the rate of the forward reaction equals the rate of the backward reaction. Setting the two rate expressions equal: (3.9e-5)[PtCl4^2-] = (2.1e-3)[Pt(H2O)Cl3^-][Cl^-]. The equilibrium constant Keq = [Pt(H2O)Cl3^-][Cl^-] / [PtCl4^2-] = 3.9e-5 / 2.1e-3 = 0.01857. The question asks for the complexation of the fourth Cl-, which is the reverse of the given reaction, so K = 1/0.01857 = 53.8.

Multiple choice chemistry carbon compounds some important carbon compounds ethanol and ethanoic acid introduction to alcohols - ethanol

Give the balance equation for the following laboratory preparation :

Ethane from sodium propionate

  1. $CH _3CH _2COONa + NaOH \rightarrow C _2H _6 + Na _2CO _3$
  2. $CH _3CH _2COONa + 2NaOH \rightarrow C _2H _4 + Na _2CO _3$
  3. $CH _3CH _2COONa + NaOH \rightarrow C _2H _4 + Na _2CO _3$
  4. $CH _3CH _2COONa + 2NaOH \rightarrow 2CH _4 + Na _2CO _3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Reaction:
$CH _3CH _2COONa + NaOH \rightarrow C _2H _6 + Na _2CO _3$

Multiple choice electrical cells patterns and properties of metals chemistry

The value of equilibrium constant for a feasible cell reaction is:

  1. $< 1$
  2. $= 1$
  3. $> 1$

  4. zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$K = antilog \left (\dfrac {nE^{\circ}}{0.0591}\right )$
For feasible cell, $E^{\circ}$ is positive, hence from the above equation $K > 1$ for feasible cell reactions.

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

The value of equilibrium constant for a feasible cell reaction must be __________.

  1. < 1

  2. Zero

  3. = 1

  4. > 1
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a feasible cell reaction, the Gibbs free energy change (Delta G) must be negative. Since Delta G = -RT ln(K), a negative Delta G implies ln(K) > 0, which means K > 1.

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

$=3\times 3=9$