Equilibrium Constant and Standard Free Energy Change

Q.

The reaction $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$ is
carried out at 298 K and 20 bar. Five moles of each of $N_2{O}_4$ and
$N{O}_2$, are taken initially. Given: $\Delta$${}_f$G${}^{\circ }$
$N_2{O}_4=100kJmo{l}^{-1}$ and $\Delta$${}_f$G${}^{\circ }$
$N{O}_2=50kJmo{l}^{-1}$. The reaction proceeds at an initial pressure
of 20 bar. Determine the direction in which the reaction proceeds to achieve
equilibrium. Also determine the amounts of $N_2{O}_4$ and $N{O}_2$ when the reaction attains equilibrium.

1. = 8.333, = 1.667

2. = 3.333, = 1.667

3. = 1.667, = 3.333

4. = 6.667, = 1.667

Q. For the reaction $2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$ The incorrect statement is :

1. Decreasing the pressure will shift the equilibrium to the left
2. Addition of inert gas at constant pressure will shift the equilibrium towards right
3. Value of $\frac{K_P}{K_C}$ is less than one
4. No change in state of equilibrium when inert gas is added at constant volume

Q. The value of $\Delta G^{\circ }$ for a reaction having K=1 would be:

1. 0
2. +RT
3. -RT
4. -1

Q. At equilibrium, reaction $A\leftrightharpoons B$ is at the stage of half completion. Which one of the followings is true?

1. 
2. 
3. 
4. 

Q. Calculate $△G^{\circ }$ for conversion of oxygen to ozone
$3/2O_2\left(g\right)\to O_3\left(g\right)at298K,if{K}_p$ for this conversion is 2.47 $\times$ ${10}^{-29}$
(Use log(2.47) = 0.4)

1. 
2. 
3. 
4. 

Q. $N_2+3H_2\rightleftharpoons 2N{H}_3$
Which is correct statement if $N_2$ is added at equilibrium condition?

1. The condition for equilibrium is $G\left(N_2\right)+3G\left(H_2\right)=2G\left(N{H}_3\right)$ where, G is Gibbs free energy per mole fo the gaseous species measured at that partial pressure. The condition of equilibrium is unaffected by the use of catalyst, which increases the rate of both the forward and backward reactions to the same extent
2. The catalyst will increase the rate fo forward reaction by $\alpha$ and that of backward reaction by $\beta$
3. The equilibrium will shift to forward direction because according to IInd law of thermodynamics, the entropy must increases in the direction of spontaneous reaction
4. Catalyst will not alter the rate fo either fo the reaction

Q. For reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by

1. With increase in temperature, the value of K for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases
2. With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases
3. With increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system is negative
4. With increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive

Q.

Thermal decomposition of gaseous $X_2$ to gaseous X at 298 K takes place according to the following equation:
$X_2\left(g\right)\rightleftharpoons 2X\left(g\right)$
The standard reaction Gibbs energy, $\Delta G^o,$ of this reaction is positive. At the start of the reaction, there is one mole of $X_2$ and no X. As the reaction proceeds, the number of moles of X formed is given by $\beta ,$ Thus, ${\beta }_{equilibrium}$ is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally.
(Given: R = 0.083 L bar $K^{-1}mo{l}^{-1}$)

The equilibrium constant $K_p$ for this reaction at 298 K, in terms of ${\beta }_{equilibrium},$ is

1. $\frac{8{\beta }_{equilibrium}^2}{4-{\beta }_{equilibrium}^2}$

2. $\frac{8{\beta }_{equilibrium}^2}{2-{\beta }_{equilibrium}}$

3. $\frac{4{\beta }_{equilibrium}^2}{2-{\beta }_{equilibrium}}$

4. $\frac{4{\beta }_{equilibrium}^2}{4-{\beta }_{equilibrium}^2}$

Q. Thermal decomposition of gaseous $X_2$ to gaseous X at 298 K takes place according to the following equation:
$X_2\left(g\right)\rightleftharpoons 2X\left(g\right)$
The standard reaction Gibbs energy, $\Delta G^o,$ of this reaction is positive. At the start of the reaction, there is one moe of $X_2$ and no X. As the reaction proceeds, the number of moles of X formed is given by $\beta ,$ Thus, ${\beta }_{equilibrium}$ is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally.
(Given: R = 0.083 L bar $K^{-1}mo{l}^{-1}$)

The Incorrect statement among the following, for this reaction is

1. $K_c<1$
2. At the start of the reaction, dissociation of gaseous $X_2$ takes place spontaneously
3. Decrease in the total pressure will result in formation of more moles of gaseous X
4. ${\beta }_{equilibrium}$ = 0.7

Q.

Rate of disappearance of the reactant A in the reversable and one step reaction $A\rightleftharpoons B$ at two temperatures is given by
$\frac{-d\left[A\right]}{dt}=2.0\times {10}^{-3}{s}^{-1}\left[A\right]-5.0\times {10}^{-4}{s}^{-1}\left[B\right]\left(at{27}^{\circ }C\right)\frac{-d\left[A\right]}{dt}=8.0\times {10}^{-2}{s}^{-1}\left[A\right]-4.0\times {10}^{-3}{s}^{-1}\left[B\right]\left(at{127}^{\circ }C\right)$
The enthapy of the reaction in the given temperature range is

1. $-\frac{2.303\times 8.314\times 300\times 400}{100}log5J$

2. $\frac{2.303\times 8.314\times 300\times 400}{100}log50J$

3. $\frac{2.303\times 8.314\times 300\times 400}{100}log5J$

4. $\frac{100}{2.303\times 8.314\times 300\times 400}log5J$