Tag: relationship between equilibrium constant, reaction quotient and gibbs energy

Questions Related to relationship between equilibrium constant, reaction quotient and gibbs energy

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.

$\Delta G$ for the reaction at 1000 K is : 

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

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

  1. $2930\;J$

  2. $2934\;J$

  3. $2300\;J$

  4. None of these

Show answer
Correct Option: A
Explanation:

A.2930J
$\triangle G=\triangle H-T\triangle S\ \triangle G=35040-1000\times 32.11\ \quad \quad \quad \quad =35040-32110\ \quad \quad \quad \quad =2930J$

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 :

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

  1. 3000 Kelvin

  2. 5000 Kelvin

  3. 5333 Kelvin

  4. 5 Kelvin

Show answer
Correct Option: B
Explanation:

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

A reaction mixture containing $H _2,\, N _2$ and $NH _3$ has partial pressure 2 atm, 1atm and 3 atm respectively at 725 K. If the value of K for the 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 relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases

  1. forward

  2. backward

  3. no net reaction

  4. direction of reaction cannot be predicted

Show answer
Correct Option: B
Explanation:

$Q _p =\dfrac{p^2 _{NH _3}}{p _{N _2}p^3 _{H _2}}$ $=\dfrac{3^2}{1 \times 2^3}=1.125 \ atm^{-1}$

Since it is grater that $K _p$ the reaction will go in backward direction to achieve $K _P$

The equilibrium constant of the reaction $2C _3H _6 (g) \rightleftharpoons C _2H _4 (g) + C _4H _8 (g)$ is found to fit the expression:
                         $lnK=-1.04-\dfrac {1088}{T}$


Calculate the standard reaction enthalpy and entropy at 400 K :

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

  1. $\Delta H^o = 4.5 kJ/mol ; \Delta S^o = 4.32 J / mol^{1} K^{1}$

  2. $\Delta H^o = 9.04 kJ/mol ; \Delta S^o =- 8.64 J / mol^{1} K^{1}$

  3. $\Delta H^o = 18.08 kJ/mol ; \Delta S^o = 17.28 J / mol^{1} K^{1}$

  4. None of these

Show answer
Correct Option: B
Explanation:
$\displaystyle  \Delta G^0 = -RTlnK =\Delta H^0 - T\Delta S^0$

$\displaystyle  lnK = \frac {\Delta S^0}{R}-\frac {\Delta H^0 }{RT} $......(1) 

But $\displaystyle lnK=-1.04-\frac {1088}{T} $......(2)

From (1) and (2),

$\displaystyle  \frac {\Delta S^0}{R} = -1.04 $ and $\displaystyle \frac {\Delta H^0 }{RT} =  \frac {1088}{T}$

$\displaystyle  \Delta S^0 = -1.04 \times 8.314 = - 8.64 J/mol/K$

$\displaystyle \Delta H^0  = 1088 \times 8.314 =9040 J/mol =9.04 kJ/mol $

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 $4\times10^{-3}$

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

  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.

Show answer
Correct Option: A

${ K } _{ P }$ 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.

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

  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. ${ K } _{ C }= 1.5$
    ${ \triangle G }^{ \oplus }= + 15.22 kcal$

Show answer
Correct Option: C
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 $K _p$

The value of $\Delta G^{\circ} _{r}$ of gaseous mercury is $31 kJ/mole$. At what external pressure mercury start boiling $25^{\circ}C$.

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

  1. $10^{-5.44}$

  2. $10^{-12.5}$

  3. $10^{-6.52}$

  4. $10^{-3.12}$

Show answer
Correct Option: A
Explanation:

The reaction occuring is
$Hg:(l)\rightleftharpoons G(g)$
$:\Delta G^{\circ}=31:kJ/mol$.
Since Boiling takes place, $K _p = P _{Hg}$.
Now, at boiling equilibrium exists. For equilibrium, we have the following equation,
$\Delta G^{\circ}=-RT:\ln :K _{P}$
$\Rightarrow 31\times 10^{3}=-8.31\times 298:\ln :K _{P}$
Solving, we get $P _{Hg} = K _p = 10^{-5.44} $

When reaction is at standard state at equilibrium, then:

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

  1. $\Delta H^\circ\, =\, 0$

  2. $\Delta S^\circ\, =\, 0$

  3. equilibrium constant $K = 0$

  4. equilibrium constant $K = 1$

Show answer
Correct Option: D
Explanation:

If $K$ is equal to 1, the reaction will reach equilibrium as an intermediate mixture, meaning the amounts of products and reactants will be about the same.

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

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

  1. < 1

  2. Zero

  3. = 1

  4. > 1

Show answer
Correct Option: D

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 :

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

  1. 9/16

  2. 1/9

  3. 16/9

  4. 9

Show answer
Correct Option: D
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 $K _c=\dfrac{[AD][CB]}{[AB][CD]}=\dfrac{\dfrac{3/4}{v}M\times \dfrac{3/4}{v}M}{\dfrac{1/4}{v}M\times \dfrac{1/4}{v}M}$

$=3\times 3=9$