Gibbs energy change and equilibrium - class-XI

At equilibrium $\Delta G=0$
$\Delta G=-nRT ln K$
$0=-nRT ln K$
or $K=1$

${ \triangle G }^{ 0 }=-RTlnK$             $T={ 27 }^{ 0 }C=300K$

The Van't Hoff equation gives the relationship between the standard gibbs free energy change and the equilibrium constant.  It is represented by the equation $-\Delta  G^{\circ} = RT    log _e   K _p$.

$\text{Option B is correct.}$

$\Delta G = 2.303RT\space logK$

As we know,
At equilibrium $\Delta G=0$

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

$K _1=\frac {K _f}{K _b}=\frac {2\times 10^{-2}}{4\times 10^{-3}}=5$ 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 $\Delta H=3.85 kcal$

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] $
$\therefore \triangle H=,12140\quad cal\quad =,-12140\quad cal$

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

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.

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.

$\Delta H^{ \ominus } = \Delta H^{ \ominus } _ { P } - \Delta H^{ \ominus } _{ R }; (for\quad 1\quad mol)$
$= \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 $5.81\quad mol\quad of\quad CaSO _4\cdot 2H _2O = 5.81\times 83.3 kJ mol^{ -1 }$
$=484 kJ mol^{ -1 }$
$\Delta G = \Delta H - T\Delta S$
$= 17.92 kJ$ ........ $(\Delta S = S _P - S _R)$
$\Delta G = -nRTln K$
$\therefore$ $K = e^{ -\Delta GlnRT } < 1$

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

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
$\Delta G^{ \ominus } = - RT log K$

As we know,

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.

At 288 K, $M _{avg.}=\frac {3.62\times 0.0821\times 288}{1}$
$\frac {92}{M _{avg.}}=1+\alpha \Rightarrow K _{P1}=\frac {4\alpha^2}{1-\alpha^2}$
Similarly at $348 K, M'/avg.=\frac {1.84\times 0.0821\times 348}{1}$
$\frac {92}{M'avg}=1+\alpha'\Rightarrow K _{P _2}=\frac {4\alpha'^2}{1-\alpha'^2}$
$log \frac {K _{P _2}}{K _{P _1}}=\frac {\Delta H^o}{2.303 R}\left [\frac {1}{288}-\frac {1}{348}\right ]$
so,
$\Delta _rH = 75.68 kJ mol^{1}$

$\Delta G = \Delta G^{ \ominus } + RT log K$
and at equilibrium,
$\Delta G = 0$ so
$\Delta G^{ \ominus } = - RT log K$
$\Delta G^{ \ominus } = -RTln K _P$
$K _P = e^{ -\Delta G^{ \ominus }/RT }$

Consider a reaction, $A+B\rightleftharpoons C+D$

The Van't Hoff equation is ln(K) = -deltaH/R * (1/T) + constant. Plotting ln(K) vs 1/T results in a straight line with a slope of -deltaH/R.

If we know $\displaystyle \Delta G^o $  of a reaction, the following can be defined. 
I. Cell potential, $\displaystyle E^o $
III. Equilibrium constant, Keq
$\displaystyle \Delta G^o = -nF E^o = - RTlnK $

 The temperature is 25 deg C or 298 K.
$\displaystyle  \Delta G^o = -RT ln K = - 8.314 \times 298 \times ln 1.3 \times 10^4 = -23469 J = -23.4 kJ$
Hence, the standard free energy change is -23.4 kJ

$\Delta G $= -2.303 RT log K
 = $-2.303 \times 8.314 \times 300 \ log 10$
=-5.74 kJ

$\Delta G^0 = 0 $ at equilibrium under standard state.
Also at equilibrium, $\Delta G  = 0 $
$\therefore$ $\Delta H^0 - T\Delta S^0 = 0$
Also $\Delta G^0 $= -2.303 RT Iog K  
$\therefore$ K = 1

$\Delta G^{o}=- R TlnKeq$

$\Delta G^{o}=$ will be negative for spontaneous process

$Keqm > 1$for spontaneous process 

$E^{o} cell > O$ for spontaneous process

$nFE _{cell}^{o}= RT ln Keq$

$\therefore E _{cell}^{o}>o$

 $\displaystyle \Delta G^o = -RTlnK _1$
 $\displaystyle -16500 = - 8.314 \times 298 \times lnK _1$
$\displaystyle 6.6597 = ln K _1$
 $\displaystyle K _1 = 779.4$
$\displaystyle K _2 = K _1^2 = (779.4)^2 = 6.074 \times 10^5$
 $\displaystyle K _3 = \dfrac {1}{K _1}=\dfrac {1}{779.4}=1.283 \times 10^{-3} $

 The temperature is 25 deg C or 298 K
$\displaystyle  \Delta G^0 = -RT lnK$
$\displaystyle  \Delta G^0 = - 107.2 kJ = -107200 J$
$\displaystyle  -107200 = - 8.314 \times 298 \times ln K$
$\displaystyle  ln K = 43.268$
$\displaystyle  K = 6.18 \times 10^{18}$

A large positive value of $\Delta { G }^{ o }$ corresponds to small positive $K$.
$\Delta { G }^{ o }=-2.303RT\log { { K } _{ c } } $
When $ \displaystyle  { K } _{ c }  >0$, $ \displaystyle \Delta { G }^{ o } <0 $ and vice versa.

If $\Delta G = 0$ this means, the system is at equilibrium at standard conditions.
$\Delta G = 0$, it means the reaction is equilibrium at standard conditions.
A negative value of $\Delta G  $, means spontaneous.
A positive value of $\Delta G $, means non-spontaneous.

The standard Gibbs free energy change for the reaction is calculated using the standard cell potential. The reaction involves oxidation of Au and reduction of O2. The expression involves the formation constant X and the standard potentials provided.

$\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}=-54.07 \times 1000 - 298 \times 10$

$=-54070-2980=-57050$

$\Delta G^{0}=-2.303 RT log _{10}K$

$-57050=-2.303\times 298\times 8.314 log _{10}K=-5705 log _{10}K$