Tag: energetics and thermochemistry

Questions Related to energetics and thermochemistry

${ K } _{ C }$ 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 ${ NH } _{ 3 }$.

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

  1. $\therefore \triangle H =\,-50462\quad cal$

  2. $\therefore \triangle H=\,-8133\quad cal$

  3. $\therefore \triangle H =\,12140\quad cal$

  4. $\therefore \triangle H=\,-12140\quad cal$

Show answer
Correct Option: D
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] $
$\therefore \triangle H=\,12140\quad cal\quad =\,-12140\quad cal$

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

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

  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 $[NO _2]=0.212 M$

Show answer
Correct Option: A,B,C
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.

The equilibrium constant is $10$ at $100 K$. Hence, $\Delta G$ will be negative.
chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

The equilibrium constant is 10 at 100 K. Hence, $\Delta G$ will be negative.
$\Delta G = -RTlnK = -RTln10 = -RT \times 2.303$ Since R and T are positive, $\Delta G$ is negative.

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

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

  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. $K _p=4\alpha^3\left [\frac {P}{1+2\alpha}\right ]^2$

Show answer
Correct Option: A,B,C,D
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.

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:

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

  1. 0

  2. <1

  3. >1

  4. =1

Show answer
Correct Option: B
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 $5.81\quad mol\quad of\quad CaSO _4\cdot 2H _2O = 5.81\times 83.3 kJ mol^{ -1 }$
$=484 kJ mol^{ -1 }$
$\Delta G = \Delta H - T\Delta S$
$= 17.92 kJ$ ........ $(\Delta S = S _P - S _R)$
$\Delta G = -nRTln K$
$\therefore$ $K = e^{ -\Delta GlnRT } < 1$

${\Delta G ^{0}}$ is related to K by the relation _____.

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

  1.  ${\Delta G ^{0}}$ =$ -RT: InK^{2}$.

  2.  ${\Delta G ^{0}}$ =$ -RTK$.

  3.  ${\Delta G ^{0}}$ =$ RT: InK$.

  4.  ${\Delta G ^{0}}$ =$ -RT: InK$.

Show answer
Correct Option: D
Explanation:
$G=G^o+RTlnK$.

$G^o$ is standard Gibbs energy. At equillibrium when there is no longer free energy left to drive the reaction then $\Delta G=0, \Delta G^o=-RTlnK$ 

The correct relationship between free energy change in a reaction and the corresponding equilibrium constant $K$ is 

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

  1. $-\Delta G=RT:\ln:K$

  2. $\Delta G^{o}=RT:\ln:K$

  3. $\Delta G=-RT:\ln:K$

  4. $-\Delta G^{o}=RT:\ln:K$

Show answer
Correct Option: C
Explanation:

The correct relationship between free energy change in a reaction and the corresponding equilibrium constant K is $-\Delta G=RT:ln:K$ or $\Delta G=-RT:ln:K$

$\Delta G = \Delta G^{ \ominus } + RT log K$
chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

As we know,
$\Delta G = \Delta H - T\Delta S$
Also, 
$\Delta G = \Delta G^{ \ominus } + RT log K$
and at equilibrium,
$\Delta G = 0$ so
$\Delta G^{ \ominus } = - RT log K$

For the reaction at $298 K$


$A (g) + B (g)\rightleftharpoons C (g) + D (g)$

$\Delta H^o = 29.8 kcal ; \Delta S^o = 0.1 kcal/K$

Calculate $\Delta G^o$ and $K$.

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

  1. $\Delta G^o = 0 ; K = 1$

  2. $\Delta G^o = 1 ; K = e$

  3. $\Delta G^o = 2 ; K = e^2$

  4. None of these

Show answer
Correct Option: A
Explanation:

As we know,


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

         $= 29.8 - ( 298\times0.1 )$

         $= 29.8-29.8=0$

Therefore, $\Delta G^o = 0$

The relation between $\Delta G^0 $ and $K$

$\Delta G^0$ = $ - RT lnK$

$K = 1 $

So, the correct option is $A$

When $\displaystyle \Delta G$ is zero :

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

  1. reaction moves in forward direction

  2. reaction moves in backward direction

  3. system is at equilibrium

  4. none of these

Show answer
Correct Option: C
Explanation:

When $\displaystyle \Delta G$ is zero, system is at equilibrium.
Positive free energy change corresponds to non spontaneous reaction.
Negative free energy change corresponds to spontaneous reaction.