Tag: gibbs energy change and equilibrium

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Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

Although dissolution of $NH _{4}Cl$ in water is endothermic yet it dissolves because:

  1. $\Delta $ G is positive
  2. $\Delta $ H is positive
  3. $\Delta $ S is positive
  4. $\Delta $ A is positive
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Although the dissolution of ammonium chloride is endothermic (Delta H > 0), it dissolves spontaneously because the increase in entropy (Delta S > 0) is sufficiently large to make the overall Gibbs free energy change negative (Delta G = Delta H - T * Delta S < 0) at room temperature.

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

The correct relationship between free energy change in a reaction and the corresponding equilibrium constant $\displaystyle { K } _{ c }$ is:

  1. $\displaystyle { \Delta G }^{ o }=RTIn{ K } _{ c }$
  2. $\displaystyle -{ \Delta G }^{ o }=RTIn{ K } _{ c }$
  3. $\displaystyle { \Delta G }=RTIn{ K } _{ c }$
  4. $\displaystyle -{ \Delta G }=RTIn{ K } _{ c }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle \because \quad \Delta G={ \Delta G }^{ o }+RTInQ$, where Q is reaction quotient at equilibrium, $\displaystyle \Delta G=0$ and $\displaystyle Q={ K } _{ c }$
$\displaystyle \therefore \quad -\Delta G=RTIn{ K } _{ c }$

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

For the reaction : $\displaystyle 2NOCl(g)\longrightarrow 2NO(g)+{ Cl } _{ 2 }(g)$, The equilibrium constant at 400K, if $\displaystyle { \Delta H }^{ o }=77.18kJ{ mol }^{ -1 }$ and $\displaystyle { \Delta S }^{ o }=0.122kJ{ K }^{ -1 }{ mol }^{ -1 }$ is:

  1. $\displaystyle 1.97\times { 10 }^{ -3 }$
  2. $\displaystyle 1.97\times { 10 }^{ -2 }$
  3. $\displaystyle 1.97\times { 10 }^{ -4 }$
  4. $\displaystyle 1.97\times { 10 }^{ -1 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given the reaction: $2NOCl(g)\rightarrow 2NO(g)+Cl _2(g)$

$\Delta G^o=\Delta H^o-T\Delta S^o$

$\Delta G^o=77.18-400\times 0.122kJmol^{-1}$

$\Delta G^o=28.38\ kJmol^{-1}$

$K=e^{(\dfrac{-\Delta G^o}{RT})} $

$=1.97\times 10^{-4}$

Hence, option C is correct.
Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

van't Hoff equation is

  1. $(d/dT) ln K=-\Delta H/RT^2$
  2. $(d/dT) ln K=+\Delta H/RT^2$
  3. $(d/dT) ln K=-\Delta H/RT$
  4. $K=Ae^{\Delta H/RT}$
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation

The van't Hoff equation provides information about the temperature dependence of the equilibrium constant. The van't Hoff equation may be derived from the Gibbs-Helmholtz equation, which gives the temperature dependence of the Gibbs free energy.
The van't Hoff equation is $K=Ae^{\Delta H/RT}$
or $\frac {d ln K}{\partial T}=\frac {\Delta H}{RT^2}$

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

The rate of disappearance of A at two temperatures is given by $A\rightleftharpoons B$
i. $\frac {-d[A]}{dt}=2\times 10^{-2}[A]-4\times 10^{-3}[B]$ at 300 K
ii. $\frac {-d[A]}{dt}=4\times 10^{-2}[A]-16\times 10^{-4}[B]$ at 300 K
From the given values of heat of reaction which are incorrect

  1. $3.86 kcal$
  2. $6.93 kcal$
  3. $1.68 kcal$
  4. $1.68\times 10^{-2} kcal$
Reveal answer Fill a bubble to check yourself
B,C,D Correct answer
Explanation

$K _1=\frac {K _f}{K _b}=\frac {2\times 10^{-2}}{4\times 10^{-3}}=5$ at 300 K
$K _2=\frac {K _f}{K _b}=\frac {4\times 10^{-2}}{16\times 10^{-4}}=25$ at 400 K
$\therefore 2.303 log\frac {25}{55}=\frac {\Delta H}{2}\times \left [\frac {400-300}{400\times 300}\right ]$
or $\Delta H=3.85 kcal$

Hence, option A is correct and others are incorrect

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

${ K } _{ C }$ for ${ 3 }/{ 2{ H } _{ 2 }+{ 1 }/{ 2{ N } _{ 2 }\rightleftharpoons  } }{ NH } _{ 3 }$ are 0.0266 and $0.0129\,{ atm }^{ -1 }\quad $ respectively, at 350$^o$C and 400$^o$C. Calculate the heat of formation of ${ NH } _{ 3 }$.

  1. $\therefore \triangle H =\,-50462\quad cal$
  2. $\therefore \triangle H=\,-8133\quad cal$
  3. $\therefore \triangle H =\,12140\quad cal$
  4. $\therefore \triangle H=\,-12140\quad cal$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As we know,
$2.303\,log[\displaystyle\frac { { K } _{ { P } _{ 2 } } }{ { K } _{ { P
} _{ 1 } } }] =\frac { \triangle H }{ R } \left[ \frac { { T } _{ 2 }-{ T
} _{ 1 } }{ { T } _{ 1 }{ T } _{ 2 } }  \right] $
$2.303\,log[\displaystyle\frac
{ 0.0129 }{ 0.0266 }] =\frac { \triangle H }{ 2 } \left[ \frac {
673-623 }{ 673\times 623 }  \right] $
$\therefore \triangle H=\,12140\quad cal\quad =\,-12140\quad cal$

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

For the equilibrium at $298$ K; $N _2O _4(g)\rightleftharpoons 2NO _2(g); G _{N _2O _4}^{\ominus}=100 kJ mol^{-1}$ and $G _{NO _2}^{\ominus}=50 kJ mol^{-1}$. If 5 mol of $N _2O _4$ and 2 moles of $NO _2$ are taken initially in one litre container than which statement are correct

  1. reaction proceeds in forward direction
  2. $K _c=1$
  3. $\Delta G=-0.55 kJ, \Delta G^{\ominus}=0$
  4. At equilibrium $[N _2O _4]=4.84 M$ and $[NO _2]=0.212 M$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

$\Delta G=\Delta G^{\ominus}+2.303 RT:log Q$

$\Delta G^{\ominus}=2\times G _{NO _2}^{\ominus}-G _{N _2O _4}^{\ominus}=2\times 50-100=0$
$\therefore \Delta G=0+2.303\times 8.314\times 10^{-3}\times 298: log \displaystyle\frac {22}{5}=0-0.55 kJ$

$\therefore \Delta G=-0.55 kJ$, i.e, reaction proceeds in forward direction
Also $\Delta G^{\ominus}=0=2.303 RT:log K \therefore K=1$
Now, $\underset {\underset {5-x}{5}}{N _2O _4}=\underset {\underset {2+2x}{2}}{2NO _2}$
$\therefore K _p=\frac {(P _{NO _2})}{(P _{N _2O _4})}=1=\frac {(2+2x)^2}{5-x}$ or  $x=0.106$


So, $[N _2O _4]=5-x=4.894M,\ [NO _2]=2+2x=2.12M$

Hence, options A, B and C are correct.

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

Which are true for the reaction: $A _2\rightleftharpoons 2C+D$?

  1. If $\Delta H=0; K _p$ increases with temperature and dissociation temperature.
  2. If $\Delta H=+ve; K _p$ increases with temperature and dissociation of $A _2$ increases.
  3. If $\Delta H=-ve; K _p$ increases with temperature and dissociation of $A _2$ decreases.
  4. $K _p=4\alpha^3\left [\frac {P}{1+2\alpha}\right ]^2$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

Initial At equilibrium
$\underset {\underset {1-\alpha}{1}}{A _2}\rightleftharpoons \underset {\underset {2\alpha}{0}}{2C}+\underset {\underset {\alpha}{0}}{D}$
$K _p=(2\alpha)^2\alpha \times \left [\frac {P}{\Delta n}\right ]^2=\frac {4\alpha^3P^2}{(1+2\alpha)^2}$
Also, as we know
2.303 log $\frac {K _2}{K _1}=\frac {\Delta H}{R}\left [\frac {T _2-T _1}{T _1T _2}\right ]$ for effect temperature on K.
If $\Delta H=+ve; K _p$ increases with temperature and dissociation of $A _2$ increases.
If $\Delta H=-ve; K _p$ decreases with temperature and dissociation of $A _2$ decreases.

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

Concrete is produced from a mixture of cement, water and small stones. Small amount of gypsum, $CaSO _4\cdot 2H _2O$ is added in cement production to improve the subsequent hardening of concrete. 
The elevated temperature during the production of cement may lead to the formation of unwanted hemihydrate $CaSO _4\cdot \frac { 1 }{ 2 }H _2O$ according to reaction.

$CaSO _4\cdot 2H _2O(s)\rightarrow CaSO _4\cdot \frac { 1 }{ 2 }H _2O(s) + \frac { 3 }{ 2 }H _2O(g)$
The $\Delta _f H^{ \ominus }$ of $CaSO _4\cdot 2H _2O(s),\ CaSO _4\frac { 1 }{ 2 }H _2O(s),\ H _2O(g)$ are $-2021.0  kJ  mol^{ -1 }$, $-1575.0  kJ  mol^{ -1 }$ and $-241.8  kJ  mol^{ -1 }$ respectively. The respective values of their standard entropies are $194.0$, $130.0$ and $188.0  J  K^{ -1 }  mol^{ -1 }.$ 
$R = 8.314  J  K^{ -1 }  mol^{ -1 } = 0.0831  L  bar  mol^{ -1 }  K^{-1}$
Answer the follwoing questions on the basis of above information.
The value of equilibrium constant for reaction is:

  1. 0

  2. <1

  3. >1

  4. =1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Delta H^{ \ominus } = \Delta H^{ \ominus } _ { P } - \Delta H^{ \ominus } _{ R }; (for\quad 1\quad mol)$
$= \left[ -1575.0 kJ mol^{ -1 }-\frac { 3 }{ 2 } \times 241.8 \right] -\left[ -2021.0 kJ mol^{ -1 } \right] $
$= +83.3 kJ mol^{ -1 }$
For 1 Kg $CaSO _4\cdot 2H _2O$
Numebr of moles $=\frac { 1000}{ 172 }$
$= 5.81$
$\therefore$ Heat change for $5.81\quad mol\quad of\quad CaSO _4\cdot 2H _2O = 5.81\times 83.3 kJ mol^{ -1 }$
$=484 kJ mol^{ -1 }$
$\Delta G = \Delta H - T\Delta S$
$= 17.92 kJ$ ........ $(\Delta S = S _P - S _R)$
$\Delta G = -nRTln K$
$\therefore$ $K = e^{ -\Delta GlnRT } < 1$