At equilibrium, the value of equilibrium constant $K$ is:

The equilibrium constants of a reaction is $73$. Calculate standard free energy change.

The Van't Hoff equation is :

Standard Gibbs Free energy change $\Delta { G }^{ o }$ for a reaction is zero. The value of equilibrium constant of the reaction will be:

if for the heterogeneous equilibrium $CaCO _{3}(s)\rightleftharpoons CaO(s)+CO _{2}(g);$ K=1 at 1 atm, the temperature is given by:

A reaction attains equilibrium, when the free energy change is

Vant Hoff's equation is ___.

Calculate the standard voltage that can be obtained from an ethane oxygen fuel cell at $25^o C$.
$C _2H _6(g) + 7/2O _2(g) \rightarrow 2CO _2(g) + 3H _2O(1); \Delta G^o = -1467 \,kJ$

Expansion of a perfect gas into vacuum is related with:

Which are correct representation at equilibrium?

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

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

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:

van't Hoff equation is

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

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

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

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

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 $188.0  J  K^{ -1 }  mol^{ -1 }.$ 

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

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

For the reaction at $298 K$

When $\displaystyle \Delta G$ is zero :

The density of an equilibrium mixture of $N _2O _4$ and $NO _2$ at 101.32 $KP _a$ is 3.62 g $dm^{3}$ at 288 K and 1.84 g $dm^{3}$ at 348 K. 

Which is not correct relationship between $\Delta G^{ \ominus }$ and equilibrium constant $K _P$

The correct relation between equilibrium constant $(K)$, standard free  energy $(\Delta {G}^{o})$ and temperature $(T)$ is:

When $\ln{K}$ is plotted against $\cfrac { 1 }{ T } $ using the Van't Hoff equation, a straight line is expected with a slope equal to:

If we know $\displaystyle { \Delta G }^{ \circ  }$ of a reaction, which of the following can be defined ?
I. Cell potential, $\displaystyle { E }^{ \circ  }$
II. Activation energy, $\displaystyle { E } _{ a }$
III. Equilibrium constant, $\displaystyle { K } _{ eq }$

For the first order reaction $A\longrightarrow B+C$, carried out at $27^0C  $ if  $ 3.8\ \times \ 10^{ -16 } \%$ of the reactant molecules exists in the activated state, the ${ E } _{ a }$ (activation energy) of the reaction is:

By which of the following relations, the equilibrium constant varies with temperature?

Calculate the Standard Free Energy Change at 25 degrees celsius given the Equilibrium constant of 1.3 x 10^4.

The cell in which the following reaction occurs:
$2Fe^{3+} _{(aq)}+2I^- _{(aq)}\rightarrow 2Fe^{2+} _{(aq)}+I _{2(s)}$ has $E^o _{cell}=0.236\ V$ at $298\ K$.
The equilibrium constant of the cell reaction is:

In dynamic equilibrium condition, the reaction on both the sides occurs at the same rate and the mass on both sides of the equilibrium does not undergo any change. This condition can be achieved only when the value of $\Delta$G is :

The equilibrium constant of a reaction is 10. What will be the value of $\Delta G^0$ at 300 K?

A reaction attains equilibrium state under standard conditions. Identify the incorrect option regarding this statement.

For a spontaneous reaction the $\Delta G$, equilibrium constant $(K _{eq})$ and $E^{0} _{cell}$ will be respectively 

For a reversible reaction, if $\Delta { G }^{ o }=0$, the equilibrium constant of the reaction should be equal to:

$\Delta G^o (298 K)$ for the reaction $\dfrac12 N _2+\dfrac32H _2\overset {K _1}{\rightleftharpoons} NH _3$ is -16.5 kJ $mol^{-1}$. The equilibrium constant $(K _1)$ at $25^oC$ & the equilibrium constant $K _2$ and $K _3$ for the following reactions are
$N _2+3H _2\overset {K _2}{\rightleftharpoons} 2NH _3$
$NH _3\overset {K _3}{\rightleftharpoons } \dfrac12N _2+\dfrac32H _2$

Calculate the equilibrium constant at 25 degrees celsius given the Standard Free Energy value of - 107.2 kJ

A large positive value of $\Delta { G }^{ o }$ corresponds to which of these?

If $\Delta G$ standard is zero, this means :

If ${E} _{cell}^{o}$ for a given reaction is negative, which gives the correct relationships for the values of $\Delta { G }^{ o }$ and ${K} _{eq.}$?

Consider the reaction of extraction of gold from its ore
$Au + 2CN^{-} (aq.) + \dfrac {1}{4}O _{2}(g) + \dfrac {1}{2}H _{2}O\rightarrow Au(CN) _{2}^{-} + OH^{-}$
Use the following data to calculate $\triangle G^{\circ}$ for the reaction
$K _{f} \left {Au(CN) _{2}^{-}\right ) = X$
$O _{2} + 2H _{2}O + 4e^{-}\rightarrow 4OH^{-}; E^{\circ} = +0.41\ volt$
$Au^{3+} + 3e^{-}\rightarrow Au; E^{\circ} = + 1.5\ volt$
$Au^{3+} + 2e^{-} \rightarrow Au^{+}; E^{\circ} = + 1.4\ volt$.

The value of $log _{10}$ K for a reaction $A\rightleftharpoons B$ is: