Q: Derive the relation between \$K_p\$ and \$K_c\$ for a general chemical equilibrium reaction.

Let the gaseous reaction is a state of equilibrium is-\$a{A}_{\left( g \right)} + b{B}_{\left( g \right)} \rightleftharpoons c{C}_{\left( g \right)} + d{D}_{\left( g \right)}\$Let \${p}_{A}, {p}_{B}, {p}_{C}\$ and \${p}_{D}\$ be the partial pressure of \$A, B, C\$ and \$D\$ repectively.Therefore,\${K}_{c} = \cfrac{{\left[ C \right]}^{c} \; {\left[ D \right]}^{d}}{{\left[ A \right]}^{a} {\left[ B \right]}^{b}} ..... \left( 1 \right)\$\${K}_{p} = \cfrac{{{p}_{C}}^{c} \; {{p}_{D}}^{d}}{{{p}_{A}}^{a} \; {{p}_{B}}^{b}} ..... \left( 2 \right)\$For an ideal gas-\$PV = nRT\$\$\Rightarrow P = \cfrac{n}{V} RT = CRT\$Whereas \$C\$ is the concentration.Therefore,\${p}_{A} = \left[ A \right] RT\$\${p}_{B} = \left[ B \right] RT\$\${p}_{C} = \left[ C \right] RT\$\${p}_{D} = \left[ D \right] RT\$Substituting the values in equation \$\left( 2 \right)\$, we have\${K}_{p} = \cfrac{{\left[ C \right]}^{c} {\left( RT \right)}^{c} \; {\left[ D \right]}^{d} {\left( RT \right)}^{d}}{{\left[ A \right]}^{a}{\left( RT \right)}^{a} {\left[ B \right]}^{b} {\left( RT \right)}^{b}}\$\$\Rightarrow {K}_{p} = \cfrac{{\left[ C \right]}^{c} \; {\left[ D \right]}^{d}}{{\left[ A \right]}^{a} {\left[ B \right]}^{b}} {\left( RT \right)}^{\left[ \left( c + d \right) - \left( a + b \right) \right]}\$\$\Rightarrow {K}_{p} = {K}_{c} {\left( RT \right)}^{\Delta{{n}_{g}}} \quad \left( \text{From } \left( 1 \right) \right]\$Here,\$\Delta{{n}_{g}} =\$ Total no. of moles of gaseous product \$-\$ Total no. of moles of gaseous reactantHence the relation between \${K}_{p}\$ and \${K}_{c}\$ is-\${K}_{p} = {K}_{c} {\left( RT \right)}^{\Delta{{n}_{g}}}\$

Q: The dissociation equilibrium of a gas \${AB}_{2}\$ can be represented as: \$2A{ B }_{ 2 }\left( g \right) \rightleftharpoons 2AB\left( g \right) +{ B }_{ 2 }\left( g \right) \$The degree of dissociation is \$x\$ and is small compared to \$1\$. The expression relating the degree of dissociation \$x\$ with equilibrium constant \${K}_{p}\$ and total pressure \$p\$ is:

Let \$P\$ be the initial pressure of \$\displaystyle AB_2\$The equilibrium pressures of \$\displaystyle AB_2,\$ \$AB\$ and \$\displaystyle B_2\$ are \$P(1-x), \$ \$xP\$ and \$0.5 xP\$ respectively. The equilibrium constant expression is \$\displaystyle K_p = \dfrac {P_{AB}^2 \ P_{B_2}}{P_{AB_2}^2}\$\$\displaystyle K_p = \dfrac {[xP]^2 \times 0.5 xP }{[P(1-x)]^2}\$\$\displaystyle K_p = P\dfrac { 0.5 x^3 }{[ (1-x)]^2}\$ Since the degree of dissociation is small compared to \$1, 1-x\$ can be approximated to 1.\$\displaystyle K_p = P \times 0.5 x^3 \$ .... (1)The total pressure is \$\displaystyle P(1-x) + Px + 0.5Px = P(1+0.5x) = p\$\$\displaystyle P = \dfrac {p}{1+0.5x} \$.....(2)Substitute (2) in (1),\$\displaystyle K_p = \dfrac {p}{1+0.5x} \times 0.5 x^3 \$Since the degree of dissociation is small compared to \$1, 1+0.5x\$ can be approximated to 1.\$\displaystyle K_p = p \times 0.5 x^3 \$\$\displaystyle x = { \left( 2{ { K }_{ p } }/{ p } \right) }^{ { 1 }/{ 3 } }\$Hence, the expression relating the degree of dissociation (x) with equilibrium constant Kp and total pressure p is \$\displaystyle x = { \left( 2{ { K }_{ p } }/{ p } \right) }^{ { 1 }/{ 3 } }\$

Q: Write the expression for the equilibrium constant, \$\displaystyle { K }_{ c }\$ for each of the following reactions.(i) \$\displaystyle 2NOCl\left( g \right) \rightleftharpoons 2NO\left( g \right) +{ Cl }_{ 2 }\left( g \right) \$(ii) \$\displaystyle { 2Cu\left( { NO }_{ 3 } \right) }_{ 2 }\left( s \right) \rightleftharpoons 2CuO\left( s \right) +4{ NO }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \$(iii) \$\displaystyle { CH }_{ 3 }{ COOC }_{ 2 }{ H }_{ 5 }\left( aq \right) +{ H }_{ 2 }O\left( 1 \right) \rightleftharpoons { CH }_{ 3 }COOH\left( aq \right) +{ C }_{ 2 }{ H }_{ 5 }OH\left( aq \right) \$(iv) \$\displaystyle { Fe }^{ 3+ }\left( aq \right) +3{ OH }^{ - }\left( aq \right) \rightleftharpoons { Fe\left( OH \right) }_{ 3 }\left( s \right) \$(v) \$\displaystyle { I }_{ 2 }\left( s \right) +5{ F }_{ 2 }\rightleftharpoons 2I{ F }_{ 5 }\$

(a) The expression for the equilibrium constant is \$\displaystyle K_c = \frac {[NO]^2[Cl_2]}{[NOCl]^2} \$.(b) The expression for the equilibrium constant is \$\displaystyle K_c = [NO_2]^4[O_2] \$.(c) The expression for the equilibrium constant is \$\displaystyle K_c = \frac {[CH_3COOH][C_2H_5OH] }{[CH_3COOC_2H_5]} \$.(d) The expression for the equilibrium constant is \$\displaystyle K_c = \frac {1}{[Fe^{3+}] [OH^-]^3} \$.(e) The expression for the equilibrium constant is \$\displaystyle K_c = \frac { [IF_5]^2}{[ F_2 ]^5} \$.

Q: \$2\$ moles of \$PCl_5\$ were heated in a closed vessel of 2 litre capacity. At equilibrium, \$40\%\$ of \$PCl_5\$ is dissociated into \$PCl_3\$ and \$Cl_2\$. The value of equilibrium constant is-

\$PCl_5\$\$PCl_3\$\$Cl_2\$Initial200Change-0.80.80.8Equilibrium1.20.80.8Molar concentration0.60.40.4The expression for the equilibrium constant is \$K_c=\frac {[PCl_3][Cl_2]} {[PCl_5]}=\frac {0.4 \times 0.4} {0.6}=0.266.\$

Q: For the reaction \$PCl_3\,(g)\,+\,Cl_2\,(g)\,\rightleftharpoons\,PCl_5\,(g)\$ the value of \$K_c\$ at \$250^oC\$ is 26. The value of \$K_p\$ at this temperature will be :

\$ K_p = K_c \times (RT )^{\Delta n}\$\$\Delta n =\$ no of moles of gaseous produt - number of moles of gaseous rectanttherefore \$\Delta n = 1-2 = -1\$R = 0.0821, T = 250 + 273.15 = 523.15 and \$K_c\$ = 26so\$ K_p = K_c \times (RT)^{-1}\$ \$K_p =\dfrac{26}{0.0821\times 523.15}=0.61\$

Q: For the reaction equilibrium \$2NOBr_{\left ( g \right )}\leftrightharpoons2NO_{\left ( g \right )} + Br_{2\left ( g \right )}\$, if \$P_{Br_{2}}= \dfrac{P}{9}\$ at equilibrium and P is the initial total pressure, then the ratio \$\dfrac{K_{P}}{P}\$ is equal to :

The equilibrium reaction is \$2NOBr_{\left ( g \right )}\leftrightharpoons 2NO_{\left ( g \right )} + Br_{2\left ( g \right )}.\$The partial pressure of bromine is \$\frac {P} {9}\$. The partial pressures of \$NOBr\$ and \$NO\$ will be \$\frac {7P} {9}\$ and \$\frac {2P} {9}\$ respectively.The expression for the equilibrium constant will be \$K_p=\dfrac {P_{NO}^2P_{Br_2}} {P_{NOBr}^2}= \dfrac {(\frac {2P} {9})^2(\frac {P} {9})} {(\frac {7P} {9})^2}\$.Thus, \$\dfrac {K_p} {P}= \dfrac {4} {441}\$.

Q: At certain temperature compound \$AB_2(g)\$ dissociates according to the reaction: \$2AB_2(g)\rightleftharpoons 2AB(g)+B_2(g)\$With the degree of dissociation \$\alpha\$, which is small compared with unity. The expression of \$K_p\$, in terms of \$\alpha\$ and initial pressure P is:

The equilibrium reaction is as shown below:\$2AB_2(g)\rightleftharpoons 2AB(g)+B_2(g)\$\$P(1-\alpha) \quad \ P \ \alpha \quad \quad \quad \ P\frac {\alpha}{2}\$Here, P is the initial pressure and \$\alpha\$ is the degree of dissociation.The expression for the equilibrium constant is \$K_p=\frac {P^2_{AB} \times P_{B_2} }{P^2_{AB_2}}\$Substitute values in the above expression.\$Kp=\frac {(P\alpha)^2\cdot (\frac {P\alpha}{2})}{(P(1-\alpha))^2}\$. But \$\alpha < < 1\$Hence, \$Kp=\frac {(P\alpha)^2\cdot (\frac {P\alpha}{2})}{(P)^2}\$ . \$\therefore Kp=\frac {P\alpha^3}{2}\$.

Q: \$\log \dfrac {K_{p}}{K_{c}}+\log RT=0\$ is true relationship for the following reaction:

For the general reaction, \$aA+bB\leftrightharpoons cC+dD\$ the relationship between two equilibrium constants is: \${ K }_{ P }={ K }_{ C }(RT)^{ \Delta n }\$where, \$\Delta n\$ (Total moles of products side) - (Total moles of reactants on the reactant side). Hence \$\Delta n\$= (d+c) - (a+b). R is the gas constant found in the ideal gas law (0.0821 liter x Atm/mole/kelvin) T is the temperature of reaction, kelvin.This we can use in a relationship for the reaction: \$log\dfrac { { K }_{ P } }{ { K }_{ C } } +logRT=0\$ \$log\dfrac { { K }_{ C }(RT)^{ \Delta n } }{ { K }_{ C } } +logRT=0\$ \$log((RT)^{ \Delta n }\times (RT)=0\$ \$log(RT)^{ \Delta n+1 }=0\$ \$(RT)^{ \Delta n+1 }=0\$R is a constant and T is the temperature of reaction, so the product cant be zero. Thats why \$\Delta n+1=0\$ and \$\Delta n=-1\$so our relationship is true for the reaction (B) \$2S{ O }_{ 2 }+{ O }_{ 2 }\leftrightharpoons 2S{ O }_{ 3 }\$Because \$\Delta n\$ for this reaction is\$\Delta n\$ =2-2-1= -1conclusion: hence option (B) is the correct answer.

Q: In which of the following equilibrium, \$K_c=K_p\$?

Kp only counts with gases molecule , while Kc only counts with aqueous solution+ gases. So here reactant and product both are gaseous that's why only reaction having equal numbers of reactants and products will have kp=kc ie.change in number of moles is equal to 0.option(a) is correct.

Q: For the reaction \$N_2O_4(g)\rightleftharpoons 2NO_2(g)\$, the degree of dissociation at equilibrium is \$0.2\$ at \$1\$ atm pressure. The equilibrium constant \$K_p\$ will be:

\$\underset {\underset {1-\alpha}{1}}{N_2O_4}\rightleftharpoons \underset {\underset {2\alpha}{0}}{2NO_2}\$[Total moles \$=1-\alpha+2\alpha=1+\alpha]\$Partial pressure \$\dfrac {1-\alpha}{1+\alpha}\times P \dfrac {2\alpha}{1+\alpha}\times P\$Given, \$\alpha=0.2, P=1\$Partial pressure of \$N_2O_4=\dfrac {1-0.2}{1+0.2}\times 1=\dfrac {0.8}{1.2}=\dfrac {2}{3}\$\$p_{NO_2}=\dfrac {2\times 0.2}{1+0.2}\times 1=\dfrac {0.4}{1.2}=\dfrac {1}{3}\$\$K_p=\dfrac {[p_{NO_2}]^2}{[p_{N_2O_4}]^2}=\dfrac {1\times 1\times 3}{3\times 3\times 2}=\dfrac {1}{6}\$Hence, the correct answer is \$C\$

Question 1

Derive the relation between $K_p$ and $K_c$ for a general chemical equilibrium reaction.

Accepted Answer

Let the gaseous reaction is a state of equilibrium is-$a{A}_{\left( g \right)} + b{B}_{\left( g \right)} \rightleftharpoons c{C}_{\left( g \right)} + d{D}_{\left( g \right)}$Let ${p}_{A}, {p}_{B}, {p}_{C}$ and ${p}_{D}$ be the partial pressure of $A, B, C$ and $D$ repectively.Therefore,${K}_{c} = \cfrac{{\left[ C \right]}^{c} \; {\left[ D \right]}^{d}}{{\left[ A \right]}^{a} {\left[ B \right]}^{b}} ..... \left( 1 \right)$${K}_{p} = \cfrac{{{p}_{C}}^{c} \; {{p}_{D}}^{d}}{{{p}_{A}}^{a} \; {{p}_{B}}^{b}} ..... \left( 2 \right)$For an ideal gas-$PV = nRT$$\Rightarrow P = \cfrac{n}{V} RT = CRT$Whereas $C$ is the concentration.Therefore,${p}_{A} = \left[ A \right] RT$${p}_{B} = \left[ B \right] RT$${p}_{C} = \left[ C \right] RT$${p}_{D} = \left[ D \right] RT$Substituting the values in equation $\left( 2 \right)$, we have${K}_{p} = \cfrac{{\left[ C \right]}^{c} {\left( RT \right)}^{c} \; {\left[ D \right]}^{d} {\left( RT \right)}^{d}}{{\left[ A \right]}^{a}{\left( RT \right)}^{a} {\left[ B \right]}^{b} {\left( RT \right)}^{b}}$$\Rightarrow {K}_{p} = \cfrac{{\left[ C \right]}^{c} \; {\left[ D \right]}^{d}}{{\left[ A \right]}^{a} {\left[ B \right]}^{b}} {\left( RT \right)}^{\left[ \left( c + d \right) - \left( a + b \right) \right]}$$\Rightarrow {K}_{p} = {K}_{c} {\left( RT \right)}^{\Delta{{n}_{g}}} \quad \left( \text{From } \left( 1 \right) \right]$Here,$\Delta{{n}_{g}} =$ Total no. of moles of gaseous product $-$ Total no. of moles of gaseous reactantHence the relation between ${K}_{p}$ and ${K}_{c}$ is-${K}_{p} = {K}_{c} {\left( RT \right)}^{\Delta{{n}_{g}}}$

Question 2

The dissociation equilibrium of a gas ${AB}_{2}$ can be represented as:                         $2A{ B }_{ 2 }\left( g \right) \rightleftharpoons 2AB\left( g \right) +{ B }_{ 2 }\left( g \right) $The degree of dissociation is $x$ and is small compared to $1$. The expression relating the degree of dissociation $x$ with equilibrium constant ${K}_{p}$ and total pressure $p$ is:

Accepted Answer

Let $P$ be the initial pressure of $\displaystyle AB_2$The equilibrium pressures of $\displaystyle AB_2,$ $AB$ and $\displaystyle B_2$ are $P(1-x), $  $xP$ and $0.5 xP$ respectively. The equilibrium constant expression is $\displaystyle  K_p = \dfrac {P_{AB}^2 \ P_{B_2}}{P_{AB_2}^2}$$\displaystyle  K_p = \dfrac {[xP]^2 \times 0.5 xP }{[P(1-x)]^2}$$\displaystyle  K_p = P\dfrac { 0.5 x^3  }{[ (1-x)]^2}$ Since the degree of dissociation is small compared to $1, 1-x$ can be approximated to 1.$\displaystyle  K_p = P \times  0.5 x^3  $ .... (1)The total pressure is $\displaystyle P(1-x) + Px + 0.5Px = P(1+0.5x) = p$$\displaystyle P = \dfrac {p}{1+0.5x} $.....(2)Substitute (2) in (1),$\displaystyle  K_p = \dfrac {p}{1+0.5x} \times   0.5 x^3  $Since the degree of dissociation is small compared to $1, 1+0.5x$ can be approximated to 1.$\displaystyle  K_p =  p \times   0.5 x^3  $$\displaystyle x = { \left( 2{ { K }_{ p } }/{ p } \right)  }^{ { 1 }/{ 3 } }$Hence, the expression relating the degree of dissociation (x) with equilibrium constant Kp and total pressure p is  $\displaystyle x = { \left( 2{ { K }_{ p } }/{ p } \right)  }^{ { 1 }/{ 3 } }$

Question 3

Write the expression for the equilibrium constant, $\displaystyle { K }_{ c }$ for each of the following reactions.(i) $\displaystyle 2NOCl\left( g \right) \rightleftharpoons 2NO\left( g \right) +{ Cl }_{ 2 }\left( g \right) $(ii) $\displaystyle { 2Cu\left( { NO }_{ 3 } \right)  }_{ 2 }\left( s \right) \rightleftharpoons 2CuO\left( s \right) +4{ NO }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) $(iii) $\displaystyle { CH }_{ 3 }{ COOC }_{ 2 }{ H }_{ 5 }\left( aq \right) +{ H }_{ 2 }O\left( 1 \right) \rightleftharpoons { CH }_{ 3 }COOH\left( aq \right) +{ C }_{ 2 }{ H }_{ 5 }OH\left( aq \right) $(iv) $\displaystyle { Fe }^{ 3+ }\left( aq \right) +3{ OH }^{ - }\left( aq \right) \rightleftharpoons { Fe\left( OH \right)  }_{ 3 }\left( s \right) $(v) $\displaystyle { I }_{ 2 }\left( s \right) +5{ F }_{ 2 }\rightleftharpoons 2I{ F }_{ 5 }$

Accepted Answer

(a) The expression for the equilibrium constant is $\displaystyle K_c = \frac {[NO]^2[Cl_2]}{[NOCl]^2} $.(b) The expression for the equilibrium constant is   $\displaystyle K_c = [NO_2]^4[O_2] $.(c) The expression for the equilibrium constant is   $\displaystyle K_c = \frac {[CH_3COOH][C_2H_5OH] }{[CH_3COOC_2H_5]} $.(d) The expression for the equilibrium constant is   $\displaystyle K_c = \frac {1}{[Fe^{3+}] [OH^-]^3} $.(e) The expression for the equilibrium constant is   $\displaystyle K_c = \frac { [IF_5]^2}{[ F_2 ]^5} $.

Question 4

$2$ moles of $PCl_5$ were heated in a closed vessel of 2 litre capacity. At equilibrium, $40\%$ of $PCl_5$ is dissociated into $PCl_3$ and $Cl_2$. The value of equilibrium constant is-

Accepted Answer

$PCl_5$$PCl_3$$Cl_2$Initial200Change-0.80.80.8Equilibrium1.20.80.8Molar concentration0.60.40.4The expression for the equilibrium constant is $K_c=\frac {[PCl_3][Cl_2]} {[PCl_5]}=\frac {0.4 \times 0.4} {0.6}=0.266.$

Question 5

For the reaction $PCl_3\,(g)\,+\,Cl_2\,(g)\,\rightleftharpoons\,PCl_5\,(g)$ the value of $K_c$ at $250^oC$ is 26. The value of $K_p$ at this temperature will be :

Accepted Answer

$ K_p = K_c \times (RT )^{\Delta n}$$\Delta n =$ no of moles of gaseous produt - number of moles of gaseous rectanttherefore $\Delta n = 1-2 = -1$R = 0.0821, T = 250 + 273.15 = 523.15 and $K_c$ = 26so$ K_p = K_c \times (RT)^{-1}$ $K_p =\dfrac{26}{0.0821\times 523.15}=0.61$

Question 6

For the reaction equilibrium $2NOBr_{\left ( g \right )}\leftrightharpoons2NO_{\left ( g \right )} + Br_{2\left ( g \right )}$, if $P_{Br_{2}}= \dfrac{P}{9}$ at equilibrium and P is the initial total pressure, then the ratio $\dfrac{K_{P}}{P}$ is equal to :

Accepted Answer

The equilibrium reaction is $2NOBr_{\left ( g \right )}\leftrightharpoons 2NO_{\left ( g \right )} + Br_{2\left ( g \right )}.$The partial pressure of bromine is $\frac {P} {9}$. The partial pressures of $NOBr$ and $NO$ will be $\frac {7P} {9}$ and $\frac {2P} {9}$ respectively.The expression for the equilibrium constant will be $K_p=\dfrac {P_{NO}^2P_{Br_2}} {P_{NOBr}^2}= \dfrac {(\frac {2P} {9})^2(\frac {P} {9})} {(\frac {7P} {9})^2}$.Thus, $\dfrac {K_p} {P}= \dfrac {4} {441}$.

Question 7

At certain temperature compound $AB_2(g)$ dissociates according to the reaction: $2AB_2(g)\rightleftharpoons 2AB(g)+B_2(g)$With the degree of dissociation $\alpha$, which is small compared with unity. The expression of $K_p$, in terms of $\alpha$ and initial pressure P is:

Accepted Answer

The equilibrium reaction is as shown below:$2AB_2(g)\rightleftharpoons 2AB(g)+B_2(g)$$P(1-\alpha)  \quad        \    P \ \alpha   \quad \quad \quad   \   P\frac {\alpha}{2}$Here, P is the initial pressure and $\alpha$ is the degree of dissociation.The expression for the equilibrium constant is $K_p=\frac {P^2_{AB} \times P_{B_2} }{P^2_{AB_2}}$Substitute values in the above expression.$Kp=\frac {(P\alpha)^2\cdot (\frac {P\alpha}{2})}{(P(1-\alpha))^2}$.     But $\alpha < < 1$Hence, $Kp=\frac {(P\alpha)^2\cdot (\frac {P\alpha}{2})}{(P)^2}$ .  $\therefore Kp=\frac {P\alpha^3}{2}$.

Question 8

$\log \dfrac {K_{p}}{K_{c}}+\log RT=0$ is true relationship for the following reaction:

Accepted Answer

For the general reaction,    $aA+bB\leftrightharpoons cC+dD$ the relationship between two equilibrium constants is: ${ K }_{ P }={ K }_{ C }(RT)^{ \Delta n }$where, $\Delta n$ (Total moles of products side) - (Total moles of reactants on the reactant side). Hence $\Delta n$= (d+c) - (a+b). R is the gas constant found in the ideal gas law (0.0821 liter x Atm/mole/kelvin) T is the temperature of reaction, kelvin.This we can use in a relationship for the reaction:        $log\dfrac { { K }_{ P } }{ { K }_{ C } } +logRT=0$        $log\dfrac { { K }_{ C }(RT)^{ \Delta n } }{ { K }_{ C } } +logRT=0$        $log((RT)^{ \Delta n }\times (RT)=0$         $log(RT)^{ \Delta n+1 }=0$         $(RT)^{ \Delta n+1 }=0$R is a constant and T is the temperature of reaction, so the product cant be zero. Thats why $\Delta n+1=0$ and $\Delta n=-1$so our relationship is true for the reaction (B) $2S{ O }_{ 2 }+{ O }_{ 2 }\leftrightharpoons 2S{ O }_{ 3 }$Because $\Delta n$ for this reaction is$\Delta n$  =2-2-1= -1conclusion: hence option (B) is the correct answer.

Question 9

In which of the following equilibrium, $K_c=K_p$?

Accepted Answer

Kp only counts with gases molecule , while Kc only counts with aqueous solution+ gases. So here reactant and product both are gaseous that's why only reaction having equal numbers of reactants and products will have kp=kc ie.change in number of moles is equal to 0.option(a) is correct.

Question 10

For the reaction $N_2O_4(g)\rightleftharpoons 2NO_2(g)$, the degree of dissociation at equilibrium is $0.2$ at $1$ atm pressure. The equilibrium constant $K_p$ will be:

Accepted Answer

$\underset {\underset {1-\alpha}{1}}{N_2O_4}\rightleftharpoons \underset {\underset {2\alpha}{0}}{2NO_2}$[Total moles $=1-\alpha+2\alpha=1+\alpha]$Partial pressure $\dfrac {1-\alpha}{1+\alpha}\times P \dfrac {2\alpha}{1+\alpha}\times P$Given, $\alpha=0.2, P=1$Partial pressure of $N_2O_4=\dfrac {1-0.2}{1+0.2}\times 1=\dfrac {0.8}{1.2}=\dfrac {2}{3}$$p_{NO_2}=\dfrac {2\times 0.2}{1+0.2}\times 1=\dfrac {0.4}{1.2}=\dfrac {1}{3}$$K_p=\dfrac {[p_{NO_2}]^2}{[p_{N_2O_4}]^2}=\dfrac {1\times 1\times 3}{3\times 3\times 2}=\dfrac {1}{6}$Hence, the correct answer is $C$

Revise with Concepts

Homogeneous Equilibrium

Heterogeneous Equilibrium

Quick Summary With Stories

Classification of Chemical Equilibrium

Derive the relation between $K_{p}$ and $K_{c}$ for a general chemical equilibrium reaction.

The dissociation equilibrium of a gas $A B_{2}$ can be represented as:

$2 A B_{2} (g) ⇌ 2 A B (g) + B_{2} (g)$

The degree of dissociation is $x$ and is small compared to $1$ . The expression relating the degree of dissociation $x$ with equilibrium constant $K_{p}$ and total pressure $p$ is:

$2$ moles of $P C l_{5}$ were heated in a closed vessel of 2 litre capacity. At equilibrium, $40 %$ of $P C l_{5}$ is dissociated into $P C l_{3}$ and $C l_{2}$ . The value of equilibrium constant is-

For the reaction $P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g)$ the value of $K_{c}$ at $25 0^{o} C$ is 26. The value of $K_{p}$ at this temperature will be :

For the reaction equilibrium $2 N O B r_{(g)} ⇋ 2 N O_{(g)} + B r_{2 (g)}$ , if $P_{B r_{2}} = \frac{P}{9}$ at equilibrium and P is the initial total pressure, then the ratio $\frac{K _{P}}{P}$ is equal to :

At certain temperature compound $A B_{2} (g)$ dissociates according to the reaction:
$2 A B_{2} (g) ⇌ 2 A B (g) + B_{2} (g)$

With the degree of dissociation $α$ , which is small compared with unity. The expression of $K_{p}$ , in terms of $α$ and initial pressure P is:

$lo g \frac{K _{p}}{K _{c}} + lo g R T = 0$ is true relationship for the following reaction:

In which of the following equilibrium, $K_{c} = K_{p}$ ?

For the reaction $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ , the degree of dissociation at equilibrium is $0.2$ at $1$ atm pressure.

The equilibrium constant $K_{p}$ will be:

Revise with Concepts

Homogeneous Equilibrium

Heterogeneous Equilibrium

Quick Summary With Stories

Classification of Chemical Equilibrium

Derive the relation between Kp​ and Kc​ for a general chemical equilibrium reaction.

The dissociation equilibrium of a gas AB2​ can be represented as: 2AB2​(g)⇌2AB(g)+B2​(g) The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation x with equilibrium constant Kp​ and total pressure p is:

2 moles of PCl5​ were heated in a closed vessel of 2 litre capacity. At equilibrium, 40% of PCl5​ is dissociated into PCl3​ and Cl2​. The value of equilibrium constant is-

For the reaction PCl3​(g)+Cl2​(g)⇌PCl5​(g) the value of Kc​ at 250oC is 26. The value of Kp​ at this temperature will be :

For the reaction equilibrium 2NOBr(g)​⇋2NO(g)​+Br2(g)​, if PBr2​​=9P​ at equilibrium and P is the initial total pressure, then the ratio PKP​​ is equal to :

At certain temperature compound AB2​(g) dissociates according to the reaction: 2AB2​(g)⇌2AB(g)+B2​(g) With the degree of dissociation α, which is small compared with unity. The expression of Kp​, in terms of α and initial pressure P is:

logKc​Kp​​+logRT=0 is true relationship for the following reaction:

In which of the following equilibrium, Kc​=Kp​?

For the reaction N2​O4​(g)⇌2NO2​(g), the degree of dissociation at equilibrium is 0.2 at 1 atm pressure. The equilibrium constant Kp​ will be: