Chemical Equilibrium | JEE Main Chemistry Chapter Wise Question Papers

Q 1 :
For the given reaction, choose the correct expression of $K_{C}$ from the following:

${Fe}_{(a q)}^{3 +} + {SCN}_{(a q)}^{-} ⇌ {(FeSCN)}_{(a q)}^{2 +}$ [2024]

$K_{C} = \frac{{[{FeSCN}^{2 +}]}^{2}}{[{Fe}^{3 +}] [{SCN}^{-}]}$
$K_{C} = \frac{[{FeSCN}^{2 +}]}{{[{Fe}^{3 +}]}^{2} {[{SCN}^{-}]}^{2}}$
$K_{C} = \frac{[{Fe}^{3 +}] [{SCN}^{-}]}{[{FeSCN}^{2 +}]}$
$K_{C} = \frac{[{FeSCN}^{2 +}]}{[{Fe}^{3 +}] [{SCN}^{-}]}$

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(4)

${Fe}^{3 +} (a q) + {SCN}^{-} (a q) ⇌ {(FeSCN)}^{2 +} (a q)$

$K_{c} = \frac{[{FeSCN}^{2 +}]}{[{Fe}^{3 +}] [{SCN}^{-}]}$

Q 2 :
The equilibrium constant for the reaction ${SO}_{3} (g) ⇌ {SO}_{2} (g) + \frac{1}{2} O_{2} (g)$ is $K_{C} = 4.9 \times 10^{- 2}$ . The value of $K_{C}$ for the reaction given below is

$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ is : [2024]

4.9
49
41.6
416

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(4)

${SO}_{3} (g) ⇌ {SO}_{2} (g) + \frac{1}{2} O_{2} (g), K_{c} = 4.9 \times 10^{- 2}$ - I

Reverse the above equation. On reversing a chemical equation, equilibrium constant gets inversed.

${SO}_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ {SO}_{3} (g), K_{c}^{1} = \frac{1}{4.9 \times 10^{- 2}}$ - II

Multiply II by 2. On multiplying an equation by $n$ , equilibrium constant gets raised by a power $n$ .

$2 {SO}_{2} (g) + O_{2} (g) ⇌ 2 {SO}_{3} (g), K_{c}^{11} = {(\frac{1}{4.9 \times 10^{- 2}})}^{2} = 416.49$

Q 3 :
At –20°C and 1 atm pressure, a cylinder is filled with equal number of $H_{2}$ , $I_{2}$ and HI molecules for the reaction $H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$ , the $K_{p}$ for the process is $x \times 10^{- 1}$ , $x =$ _______ . [Given : R = 0.082 L atm $K^{- 1}$ ${mol}^{- 1}$ ] [2024]

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10
0.01

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(3)

$n_{HI} = n_{H_{2}} = n_{I_{2}} = x$

$H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$

$K_{p} = K_{c} = \frac{{[HI]}^{2}}{[H_{2}] [I_{2}]} = \frac{{(n_{HI} / V)}^{2}}{(n_{H_{2}} / V) (n_{I_{2}} / V)} = \frac{{(n_{HI})}^{2}}{n_{H_{2}} \times n_{I_{2}}} = \frac{{(x)}^{2}}{x \times x} = 10 \times 10^{- 1}$

Q 4 :
The ratio $\frac{K_{P}}{K_{C}}$ for the reaction:

${CO}_{(g)} + \frac{1}{2} O_{2 (g)} ⇌ {CO}_{2 (g)}$ is : [2024]

$R T$
${(R T)}^{1 / 2}$
$\frac{1}{\sqrt{R T}}$
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(3)

$CO (g) + \frac{1}{2} O_{2} (g) ⇌ {CO}_{2} (g)$

$Δ n_{g} = Gaseous mol of product - gaseous mol of reactant$

$= 1 - (1 + 0.5) = - 0.5$

$K_{p} = K_{c} {(R T)}^{Δ n_{g}}$

$\frac{K_{p}}{K_{c}} = {(R T)}^{- 0.5} = \frac{1}{\sqrt{R T}}$

Q 5 :
For the given hypothetical reactions, the equilibrium constants are as follows:

$X ⇌ Y; K_{1} = 1.0$

$Y ⇌ Z; K_{2} = 2.0$

$Z ⇌ W; K_{3} = 4.0$

The equilibrium constant for the reaction $X ⇌ W$ is [2024]

12.0
6.0
8.0
7.0

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(3)

On adding reactions, equilibrium constants are multiplied

$X ⇌ Y, K_{1} = \frac{[Y]}{[X]}$

$Y ⇌ Z, K_{2} = \frac{[Z]}{[Y]}$

$Z ⇌ W, K_{3} = \frac{[W]}{[Z]}$

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Adding all three:

$X ⇌ W, K = \frac{[W]}{[X]} = \frac{[W] [Z] [Y]}{[Z] [Y] [X]} = K_{1} K_{2} K_{3} = 1 \times 2 \times 4 = 8$

Q 6 :
For the reaction $N_{2} O_{4 (g)} ⇄ 2 {NO}_{2 (g)},$ $K_{p} = 0.492$ atm at 300 K. $K_{c}$ for the reaction at same temperature is _______ $\times 10^{- 2}$ . (Given: R = 0.082 L atm ${mol}^{- 1}$ $K^{- 1}$ ) [2024]

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(2)

$N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g)$

$Δ n_{g} =$ gaseous moles of products - gaseous moles of reactants

$= 2 - 1 = 1$

$K_{p} = K_{c} {(R T)}^{Δ n_{g}}$

$0.492 atm = K_{c} \times {(0.082 atm L {mol}^{- 1} K^{- 1} \times 300 K)}^{1}$

$K_{c} = \frac{0.492}{0.082 \times 300} mol L^{- 1}$

$= 0.02 mol L^{- 1} = 2 \times 10^{- 2} mol L^{- 1}$

Q 7 :
The following concentrations were observed at 500 K for the formation of ${NH}_{3}$ from $N_{2}$ and $H_{2}$ . At equilibrium; $[N_{2}] = 2 \times 10^{- 2} M, [H_{2}] = 3 \times 10^{- 2} M and [{NH}_{3}] = 1.5 \times 10^{- 2} M .$ Equilibrium constant for the reaction is _____________ . [2024]

0%

(417)

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$

$K_{c} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}} = \frac{{(1.5 \times 10^{- 2})}^{2}}{[2 \times 10^{- 2}] {[3 \times 10^{- 2}]}^{3}} = 417$

Q 8 :
$A_{(g)} ⇌ B_{(g)} + \frac{C}{2} (g)$ The correct relationship between $K_{p},$ $α$ and equilibrium pressure P is [2024]

$K_{P} = \frac{α^{1 / 2} P^{1 / 2}}{{(2 + α)}^{3 / 2}}$
$K_{P} = \frac{α^{1 / 2} P^{3 / 2}}{{(2 + α)}^{3 / 2}}$
$K_{P} = \frac{α^{3 / 2} P^{1 / 2}}{{(2 + α)}^{1 / 2} (1 - α)}$
$K_{P} = \frac{α^{1 / 2} P^{1 / 2}}{{(2 + α)}^{1 / 2}}$

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(3)

$K_{p} = \frac{P_{C}^{1 / 2} \times P_{B}}{P_{A}}$

$= \frac{{(\frac{α P}{2 + α})}^{1 / 2} \times (\frac{2 α}{2 + α}) P}{(\frac{2 - 2 α}{2 + α}) P} = \frac{α^{3 / 2} P^{1 / 2}}{{(2 + α)}^{1 / 2} (1 - α)}$

Q 9 :
A vessel at 1000 K contains ${CO}_{2}$ with a pressure of 0.5 atm. Some of ${CO}_{2}$ is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm, then $K_{p}$ is: [2025]

3 atm
1.8 atm
0.18 atm
0.3 atm

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(2)

$\begin{matrix} {CO}_{2} (g) & + & C (s) & ⟶ & 2 CO (g) \\ t = 0 & 0.5 atm & 0 & 0 \\ t = \infty & (0.5 - p) atm & 0 & 2 p atm \end{matrix}$

$Total pressure at equilibrium = P_{{CO}_{2}} + P_{CO} = 0.5 - p + 2 p = 0.5 + p = 0.8,$ $p = 0.3 atm$

Hence, $P_{{CO}_{2}} = 0.5 - p = (0.5 - 0.3) = 0.2 atm$

$P_{CO} = 2 p = 2 \times 0.3 = 0.6 atm$

$K_{p} = \frac{P_{CO}^{2}}{P_{{CO}_{2}}} = \frac{{(0.6)}^{2}}{0.2} = 1.8$

Q 10 :
For a reaction $N_{2} O_{5} (g) ⟶ 2 {NO}_{2} (g) + \frac{1}{2} O_{2} (g)$ in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is [2025]

7/4 times of initial pressure
5/2 times of initial pressure
5 times of initial pressure
7/2 times of initial pressure

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(1)

$N_{2} O_{5} (g) ⟶ 2 {NO}_{2} (g) + \frac{1}{2} O_{2} (g)$

$\begin{matrix} t = 0 & p & 0 & 0 \\ t = \infty & p - 0.5 p & 2 \times 0.5 p & \frac{1}{2} \times 0.5 p \\ = 0.5 p & = p & = 0.25 p \end{matrix} \begin{matrix} Total = 1.75 p \\ = \frac{7}{4} p \end{matrix}$

Topic Question Set

Q 1 :
For the given reaction, choose the correct expression of $K_{C}$ from the following:

${Fe}_{(a q)}^{3 +} + {SCN}_{(a q)}^{-} ⇌ {(FeSCN)}_{(a q)}^{2 +}$ [2024]

Q 2 :
The equilibrium constant for the reaction ${SO}_{3} (g) ⇌ {SO}_{2} (g) + \frac{1}{2} O_{2} (g)$ is $K_{C} = 4.9 \times 10^{- 2}$ . The value of $K_{C}$ for the reaction given below is

$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ is : [2024]

Q 3 :
At –20°C and 1 atm pressure, a cylinder is filled with equal number of $H_{2}$ , $I_{2}$ and HI molecules for the reaction $H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$ , the $K_{p}$ for the process is $x \times 10^{- 1}$ , $x =$ _______ . [Given : R = 0.082 L atm $K^{- 1}$ ${mol}^{- 1}$ ] [2024]

Q 4 :
The ratio $\frac{K_{P}}{K_{C}}$ for the reaction:

${CO}_{(g)} + \frac{1}{2} O_{2 (g)} ⇌ {CO}_{2 (g)}$ is : [2024]

Q 5 :
For the given hypothetical reactions, the equilibrium constants are as follows:

$X ⇌ Y; K_{1} = 1.0$

$Y ⇌ Z; K_{2} = 2.0$

$Z ⇌ W; K_{3} = 4.0$

The equilibrium constant for the reaction $X ⇌ W$ is [2024]

Q 6 :
For the reaction $N_{2} O_{4 (g)} ⇄ 2 {NO}_{2 (g)},$ $K_{p} = 0.492$ atm at 300 K. $K_{c}$ for the reaction at same temperature is _______ $\times 10^{- 2}$ . (Given: R = 0.082 L atm ${mol}^{- 1}$ $K^{- 1}$ ) [2024]

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Q 7 :
The following concentrations were observed at 500 K for the formation of ${NH}_{3}$ from $N_{2}$ and $H_{2}$ . At equilibrium; $[N_{2}] = 2 \times 10^{- 2} M, [H_{2}] = 3 \times 10^{- 2} M and [{NH}_{3}] = 1.5 \times 10^{- 2} M .$ Equilibrium constant for the reaction is _____________ . [2024]

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Q 8 :
$A_{(g)} ⇌ B_{(g)} + \frac{C}{2} (g)$ The correct relationship between $K_{p},$ $α$ and equilibrium pressure P is [2024]

Q 9 :
A vessel at 1000 K contains ${CO}_{2}$ with a pressure of 0.5 atm. Some of ${CO}_{2}$ is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm, then $K_{p}$ is: [2025]

Q 10 :
For a reaction $N_{2} O_{5} (g) ⟶ 2 {NO}_{2} (g) + \frac{1}{2} O_{2} (g)$ in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is [2025]

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Topic Question Set

Q 1 : For the given reaction, choose the correct expression of KC from the following: Fe(aq)3++SCN(aq)-⇌(FeSCN)(aq)2+ [2024]

Q 2 : The equilibrium constant for the reaction SO3(g)⇌SO2(g)+12O2(g) is KC=4.9×10-2. The value of KC for the reaction given below is 2SO2(g)+O2(g)⇌2SO3(g) is : [2024]

Q 3 : At –20°C and 1 atm pressure, a cylinder is filled with equal number of H2, I2 and HI molecules for the reaction H2(g)+I2(g)⇌2HI(g), the Kp for the process is x×10-1, x= _______ . [Given : R = 0.082 L atm K-1 mol-1] [2024]

Q 4 : The ratio KPKC for the reaction: CO(g)+12O2(g)⇌CO2(g) is : [2024]

Q 5 : For the given hypothetical reactions, the equilibrium constants are as follows: X⇌Y; K1=1.0 Y⇌Z; K2=2.0 Z⇌W; K3=4.0 The equilibrium constant for the reaction X⇌W is [2024]

Q 6 : For the reaction N2O4(g)⇄2NO2(g), Kp=0.492 atm at 300 K. Kc for the reaction at same temperature is _______ ×10-2. (Given: R = 0.082 L atm mol-1 K-1) [2024]

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Q 7 : The following concentrations were observed at 500 K for the formation of NH3 from N2 and H2. At equilibrium; [N2]=2×10-2 M, [H2]=3×10-2 M and [NH3]=1.5×10-2 M. Equilibrium constant for the reaction is _____________ . [2024]

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Q 8 : A(g)⇌B(g)+C2(g) The correct relationship between Kp, α and equilibrium pressure P is [2024]

Q 9 : A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of CO2 is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm, then Kp is: [2025]

Q 10 : For a reaction N2O5(g) ⟶ 2NO2(g)+12O2(g) in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is [2025]

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Q 1 :
For the given reaction, choose the correct expression of $K_{C}$ from the following:

${Fe}_{(a q)}^{3 +} + {SCN}_{(a q)}^{-} ⇌ {(FeSCN)}_{(a q)}^{2 +}$ [2024]

Q 2 :
The equilibrium constant for the reaction ${SO}_{3} (g) ⇌ {SO}_{2} (g) + \frac{1}{2} O_{2} (g)$ is $K_{C} = 4.9 \times 10^{- 2}$ . The value of $K_{C}$ for the reaction given below is

$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ is : [2024]

Q 3 :
At –20°C and 1 atm pressure, a cylinder is filled with equal number of $H_{2}$ , $I_{2}$ and HI molecules for the reaction $H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$ , the $K_{p}$ for the process is $x \times 10^{- 1}$ , $x =$ _______ . [Given : R = 0.082 L atm $K^{- 1}$ ${mol}^{- 1}$ ] [2024]

Q 4 :
The ratio $\frac{K_{P}}{K_{C}}$ for the reaction:

${CO}_{(g)} + \frac{1}{2} O_{2 (g)} ⇌ {CO}_{2 (g)}$ is : [2024]

Q 5 :
For the given hypothetical reactions, the equilibrium constants are as follows:

$X ⇌ Y; K_{1} = 1.0$

$Y ⇌ Z; K_{2} = 2.0$

$Z ⇌ W; K_{3} = 4.0$

The equilibrium constant for the reaction $X ⇌ W$ is [2024]

Q 6 :
For the reaction $N_{2} O_{4 (g)} ⇄ 2 {NO}_{2 (g)},$ $K_{p} = 0.492$ atm at 300 K. $K_{c}$ for the reaction at same temperature is _______ $\times 10^{- 2}$ . (Given: R = 0.082 L atm ${mol}^{- 1}$ $K^{- 1}$ ) [2024]

Q 8 :
$A_{(g)} ⇌ B_{(g)} + \frac{C}{2} (g)$ The correct relationship between $K_{p},$ $α$ and equilibrium pressure P is [2024]

Q 9 :
A vessel at 1000 K contains ${CO}_{2}$ with a pressure of 0.5 atm. Some of ${CO}_{2}$ is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm, then $K_{p}$ is: [2025]

Q 10 :
For a reaction $N_{2} O_{5} (g) ⟶ 2 {NO}_{2} (g) + \frac{1}{2} O_{2} (g)$ in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is [2025]