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

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

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.

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

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

  1. $2930\;J$
  2. $2934\;J$
  3. $2300\;J$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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:

  1. forward

  2. backward

  3. no net reaction

  4. direction of reaction cannot be predicted
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $K _P$

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 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 :

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $
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 $\Delta G^{\circ} _{r}$ of gaseous mercury is $31 kJ/mole$. At what external pressure mercury start boiling $25^{\circ}C$.

  1. $10^{-5.44}$
  2. $10^{-12.5}$
  3. $10^{-6.52}$
  4. $10^{-3.12}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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} $

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

When reaction is at standard state at equilibrium, then:

  1. $\Delta H^\circ\, =\, 0$
  2. $\Delta S^\circ\, =\, 0$
  3. equilibrium constant $K = 0$
  4. equilibrium constant $K = 1$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

$=3\times 3=9$