# Equilibrium

## Chemistry Class 11

### NCERT

1   A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.$\\$ (a) What is the initial effect of the change on vapour pressure?$\\$ (b) How do rates of evaporation and condensation change initially?$\\$ (c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

##### Solution :

(a) If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.$\\$ (b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially.$\\$ (c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.

2   What is $K c$ for the following equilibrium when the equilibrium concentration of each substance is:$[SO_2]=0.60 M,[O_2]=0.82M$ and $[SO_3]=1.90 M?$$\\ 2SO_2(g)+O_2(g)\leftrightarrow 2SO_3(g)$$\\$

The relation between $K _p$ and $K _c$ is given as:$\\$ $K_ p = K _c ( RT ) \Delta n$$\\ (a) Here, \Delta n = 3 - 2 = 1$$\\$ $R = 0.0831$ barLmol$^{-1} K ^{-1}$$\\ T = 500 K$$\\$ $K_ p = 1.8 × 10 -2$$\\ Now,\\ K _p = K _c ( RT ) ^{\Delta n}$$\\$ $\Rightarrow 1.8*10^{-2}=K_c(0.0831*500)^{1}$$\\ \Rightarrow K_c=\dfrac{1.8*10^{-2}}{0.0831 *500}$$\\$ $=4.33*10^{-4}$(approximately)$\\$ (b)Here,$\\$ $\Delta n=2-1=1$$\\ R=0.0831 barLmol^{-1}K^{-1}T$$\\$ $=1073 K$$\\ K_p=167$$\\$ Now,$\\$ $K_p=K_c(RT)^{\Delta n}\\ \Rightarrow 167=K_c(0.0831*1073)^{\Delta n}\\ \Rightarrow K_c=\dfrac{167}{0.0831*1073}$$\\ =1.87(approximately) 6 For the following equilibrium, K_ c = 6.3 * 10 ^{14} at 1000K \\ NO ( g ) + O _3 ( g ) \leftrightarrow NO_ 2 ( g ) + O _2 ( g ) \\ Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K _c , for the reverse reaction? ##### Solution : It is given that K _c for the forward reaction is 6.3 *10 ^{14} \\ Then, K_c for the reverse reaction will be, K '_ C =\dfrac{1}{K_C} \\ =\dfrac{1}{6.3*10^{14}} \\ =1.59*10^{-15} 7 Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression? ##### Solution : For a pure substance (both solids and liquids),\\ [Pure substance]=\dfrac{\text{Number of moles}}{\text{Volume}} \\ =\dfrac{\text{Mass/molecular mass}}{\text{Volume}} \\ =\dfrac{\text{Mass}}{\text{Volume * Molecular mass}} \\ =\dfrac{\text{Density}}{\text{Molecular mass }} \\ Now, the molecular mass and density (at a particular temperature) of a pure substance is always fixed and is accounted for in the equilibrium constant. Therefore, the values of pure substances are not mentioned in the equilibrium constant expression. 8 Reaction between N _2 and O _2 takes place as follows:\\ 2 N _2 ( g ) + O_ 2 ( g ) \leftrightarrow 2 N _2 O ( g ) \\ If a mixture of 0.482 mol of N _2 and 0.933 mol of O _2 is placed in a 10 L reaction vessel and allowed to form N _2 O at a temperature for which K _c = 2.0 * 10 ^{-37} , determine the composition of equilibrium mixture. ##### Solution : Let the concentration of N _2 O at equilibrium be x. The given reaction is:\\ 2 N _2 ( g ) + O_ 2 ( g ) \leftrightarrow 2 N _2 O ( g ) \\ Initial conc. \quad 0.482 mol \quad 0.933 mol \qquad 0$$\\$ At equilibrium $(0.482-x)$ mol $\quad (1.933-x)$mol $\quad x$ mol $\\$ Therefore ,at equlibrium,in the $10L$ vessel: $\\$ $[N_2]=\dfrac{0.482-x}{10},[O_2]=\dfrac{0.933-x/2}{10},[N_2O]=\dfrac{x}{10}$ $\\$ The value of equilibrium constant i.e. $K _c = 2.0 *10^{-37}$ is very small. Therefore, the amount of $N _2$ and $O_ 2$ reacted is also very small. Thus, $x$ can be neglected from the expressions of molar concentrations of $N _2$ and $O _2$ . Then,$\\$ $[N_2]=\dfrac{0.482}{10}=0.0482$ mol L$6{-1}$ and $[O_2]=\dfrac{0.933}{10}=0.0933$ mol L$^{-1}$ $\\$ Now $\\$ $K_C=\dfrac{[N_2O(g)]^2}{[N_2(g)][O_2(g)]}$ $\\$ $\Rightarrow 2.0*10^{-37}=\dfrac{(\dfrac{x}{10})^2}{(0.0482)^2(0.0933)}$ $\\$ $\Rightarrow \dfrac{x^2}{100}=2.0*10^{-37}*(0.0482)^2(0.0933)$ $\\$ $\Rightarrow x^2=43.35*10^{-40}$ $\\$ $\Rightarrow x=6.6*10^{-20}$ $\\$ $[N_2O]=\dfrac{x}{10}=\dfrac{6.6*10^{-20}}{10}=6.6*10^{-21}$

9   Nitric oxide reacts with $Br_ 2$ and gives nitrosyl bromide as per reaction given below:$\\$ $2 NO ( g ) + Br_ 2 ( g ) \leftrightarrow 2 NOBr ( g )$ $\\$ When $0.087$ mol of $NO$ and $0.0437$ mol of $Br _2$ are mixed in a closed container at constant temperature, $0.0518$ mol of $NOBr$ is obtained at equilibrium. Calculate equilibrium amount of $NO$ and $Br_ 2 .$

##### Solution :

The given reaction is:$\\$ $2NO (g) +Br _2(g) \leftrightarrow 2NOBr (g)$ $\\$ $2$mol $\quad 1$mol$\quad 2$mol $\\$ Now, $2$ mol of $NOBr$ are formed from $2$ mol of $NO$. Therefore, $0.0518$ mol of $NOBr$are formed from $0.0518$ mol of $NO.$ $\\$ Again,$2$ mol of $NOBr$ are formed from $1$ mol of Br.$\\$ Therefore, $0.0518$ mol of $NOBr$ are formed from $\dfrac{0.0518}{2}$ mol of $Br$, or $0.0259$ mol of $NO.$ $\\$ The amount of $NO$ and $Br$ present initially is as follows:$\\$ $[NO] = 0.087$ mol $[Br_ 2 ] = 0.0437$ mol$\\$ Therefore, the amount of $NO$ present at equilibrium is:$\\$ $[NO] = 0.087 - 0.0518 = 0.0352$ mol$\\$ And, the amount of $Br$ present at equilibrium is:$\\$ $[Br_ 2 ] = 0.0437 - 0.0259 = 0.0178$ mol $\\$

10   At $450 K, K _p = 2.0 *10 ^{10} /$bar for the given reaction at equilibrium.$\\$ $2 SO _2( g ) + O_ 2( g ) \leftrightarrow 2 SO _3( g )$ $\\$ What is $K_ c$ at this temperature?

##### Solution :

For the given reaction,$\\$ $\Delta n = 2 - 3 = - 1$ $\\$ $T = 450 K$ $\\$ $R = 0.0831$ bar $L$ bar$K ^{-1 }$mol $^{-1}$ $\\$ $K _p =2.0 * 10 ^{10}$ bar$^{-1}$ $\\$ We know that,$\\$ $K_ P + K_ C ( RT ) \Delta n$ $\\$ $\Rightarrow 2.0 * 10^{ 10}$ bar $^{-1 }=K _C (0.0831 L$ bar $K ^{-1 }$mol $^{- 1 }* 450 K )^{- 1}$ $\\$ $\Rightarrow K_C=\dfrac{2.0*10^{10} \ bar ^{-1}}{(0.0831 L \ bar ^{-1} \ mol ^{-1}* 450 K)^{-1}}$ $\\$ $=(2.0*10^{10} \ bar^{-1})(0.0831 L \ bar \ K^{-1} \ mol ^ {-1} * 450 K)$ $\\$ $=74.79*10^{14}L$ mol$^{1}$