For the following three reactions (i), (ii) and (iii), equilibrium constants are given
(i) $ {CO_{(g)} + H_2O_{(g)} <=> CO_{2(g)} + H_{2(g)} ; K_1}$
(ii) $ { CH_{4(g)} + H_2O_{(g)} <=> CO_{(g)} 3H_{2(g)} ; K_2}$
(iii) $ {CH_{4(g)} + 2H_2O_{(g)} <=> CO_{2(g)} + 4H_{2(g)} ; K _3}$ Which of the following relation is correct?
- $K_3 . L_2^3 = K_1^2 $
- $K_1 \sqrt{K_2} = K_3$
- $K_2K_3 = K_1$
- $K_3 = K_1 K_2$
The Correct Option is D
Solution and Explanation
(i) $ {CO{(g)} + H_2O{(g)} <=> CO_2{(g)} + H_{2(g)} ; K_1}$
Equilibrium constant $\left(K_{1}\right)=\frac{[CO_{2}]\left[ H2\right]}{\left[CO\right]\left[H_{2}O\right]}$
(ii) $ { CH_{4(g)} + H_2O{(g)} <=> CO{(g)} 3H_{2(g)} ; K_2}$
Equilibrium constant $\left(K_{2}\right)=\frac{\left[CO\right]\left[H_{2}\right]^{3}}{\left[CH_{4}\right]\left[H_{2}O\right]}$
(iii) $ {CH_{4(g)} + 2H_2O_{(g)} <=> CO_{2(g)} + 4H_{2(g)} ; K _3}$
Equilibrium constant $\left(K_{3}\right)=\frac{\left[CO_{2}\right]\left[H_{2}\right]^{4}}{\left[CH_{4}\right]\left[H_{2}O\right]^{2}}$
So,$ K_{3}=K_{1}.K_{2}$
Top Questions on Equilibrium Constant
- In which of the following equilibria, Kp and Ke are NOT equal?
- NEET (UG) - 2024
- Chemistry
- Equilibrium Constant
- At \(298\ K\)
\(N_2( g)+3H_2( g)⇌2NH_3( g),K_1=4×10^5\)
\(N_2( g)+O_2( g)⇌2NO(g),K_2=1.6×10^{12}\)
\(H_2( g)+1 2O_2( g)⇌H_2O(g),K_3=1.0×10^{−13}\)
Based on above equilibria, the equilibrium constant of the reaction,
\(2NH_3( g)\)+\(\frac5 {2}\) \(O_2( g)⇌2NO(g)+3H_2O(g)\) is _______ \(×10^{−33 }\). (Nearest integer)- JEE Main - 2023
- Chemistry
- Equilibrium Constant
- The equilibrium constant for the reaction
\(Zn(s)+Sn^{2+}(aq)⇌Zn^{2+}(aq)+Sn(s)\, is\, 1×10^{20}\) at \(298 K\).
The magnitude of standard electrode potential of \(Sn/Sn^{2+}\) if \(E^{\degree}_{ Zn^{2+}/Zn}= –0.76V \) is _______ \(×10^{−2} V\). (Nearest integer).
(Given : \(\frac{2.303RT}{ F}\) \(=0.059 V \))- JEE Main - 2023
- Chemistry
- Equilibrium Constant
- At \(298\ K\), the equilibrium constant is \(2 × 10^{15}\) for the reaction:
\(Cu(s)+2Ag+(Aq)⇌Cu^{2+}(aq)+2Ag(s)\)
The equilibrium constant for the reaction:
\(\frac 12Cu^{2+}(aq)+Ag(s)+(Aq)⇌\frac 12Cu(s)+Ag^+(aq)\) is \(x \times10^{-8}\).
The value of \(x\) is ______ . (Nearest integer)- JEE Main - 2022
- Chemistry
- Equilibrium Constant
- \(PCl_5\) dissociates as
\(PCl_5(g) ⇌ PCl_3(g)+ Cl_2(g)\)
\(5\) moles of \(PCl_5\) are placed in a \(200\) litre vessel which contains \(2\) moles of \(N_2\) and is maintained at \(600\ K\). The equilibrium pressure is \(2.46\) atm. The equilibrium constant Kp for the dissociation of \(PCl_5\) is ______ \(× 10^{–3}\). (nearest integer)
(Given: \(R = 0.082\) L atm K–1 mol–1; Assume ideal gas behaviour)- JEE Main - 2022
- Chemistry
- Equilibrium Constant
Questions Asked in UPSEE exam
- A carnot engine takes 300 calories of heat at 500 K and rejects 150 calories of heat to the sink. The temperature of the sink is
- UPSEE - 2019
- Heat Transfer
- Which of the following is not permissible arrangement of electrons in an atom.
- UPSEE - 2019
- Quantum Mechanical Model of Atom
- If det $\begin{bmatrix}1&1&2\\ 2&4&9\\ t&t^{2}&1+t^{3}\end{bmatrix} = 0 $ , then the values of t are
- UPSEE - 2019
- Properties of Determinants
- The imaginary part of $\left( \frac{1}{2} + \frac{1}{2}i\right)^{10} $ is
- UPSEE - 2019
- Complex Numbers and Quadratic Equations
- Let S be the set of all right angled triangles with integer sides forming consecutive terms of an arithmetic progression. The number of triangles in S with perimeter less than 30 is
- UPSEE - 2019
- Arithmetic Progression
Concepts Used:
Equilibrium Constant
The equilibrium constant may be defined as the ratio between the product of the molar concentrations of the products to that of the product of the molar concentrations of the reactants with each concentration term raised to a power equal to the stoichiometric coefficient in the balanced chemical reaction.
The equilibrium constant at a given temperature is the ratio of the rate constant of forwarding and backward reactions.
Equilibrium Constant Formula:
Kequ = kf/kb = [C]c [D]d/[A]a [B]b = Kc
where Kc, indicates the equilibrium constant measured in moles per litre.
For reactions involving gases: The equilibrium constant formula, in terms of partial pressure will be:
Kequ = kf/kb = [[pC]c [pD]d]/[[pA]a [pB]b] = Kp
Where Kp indicates the equilibrium constant formula in terms of partial pressures.
- Larger Kc/Kp values indicate higher product formation and higher percentage conversion.
- Lower Kc/Kp values indicate lower product formation and lower percentage conversion.
Medium Kc/Kp values indicate optimum product formation.
Units of Equilibrium Constant:
The equilibrium constant is the ratio of the concentrations raised to the stoichiometric coefficients. Therefore, the unit of the equilibrium constant = [Mole L-1]△n.
where, ∆n = sum of stoichiometric coefficients of products – a sum of stoichiometric coefficients of reactants.