The equilibrium constant for the following reactions $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), N_2(g) +O_2(g) \rightleftharpoons 2NO(g)$ and$ H_2(g) + \frac{1}{2} O_2(g) \rightleftharpoons H_2O(1g)$ are K1, K2 and K3 respectively. The equilibrium constant (K) for the reaction
- $K_2.K_3^ 3/K_1$
- $K_2^ 2K_3/K_1$
- $K_1.K_2/K_3 ^2$
- $K_2.K_3/K_1 ^2$
The Correct Option is A
Approach Solution - 1
Approach Solution -2
Given equilibrium constants:
\(K_1\ for\ \text{N}_2 \,(\text{g}) + 3\text{H}_2 \,(\text{g}) \rightleftharpoons 2\text{NH}_3 \,(\text{g})\)
\(K_2 \ for \ \text{N}_2 \,(\text{g}) + \text{O}_2 \,(\text{g}) \rightleftharpoons 2\text{NO} \,(\text{g})\)
\(K_3 \ for \ \text{H}_2 \,(\text{g}) + \frac{1}{2}\text{O}_2 \,(\text{g}) \rightleftharpoons \text{H}_2\text{O} \,(\text{g})\)
We need to find the equilibrium constant K for the reaction:
\(2\text{NH}_3 \,(\text{g}) + \frac{5}{2}\text{O}_2 \,(\text{g}) \rightleftharpoons 2\text{NO} \,(\text{g}) + 3\text{H}_2\text{O} \,(\text{g})\)
Multiplying equation (iii) by 3, we get equation (iv):
\(3\text{H}_2 \,(\text{g}) + \frac{3}{2}\text{O}_2 \,(\text{g}) \rightleftharpoons 3\text{H}_2\text{O} \,(\text{g}); \quad (K_3)^3\)
Adding equations (i), (ii), and (iv), we get equation (v):
\(2\text{NH}_3 \,(\text{g}) + \frac{5}{2}\text{O}_2 \,(\text{g}) \rightleftharpoons 2\text{NO} \,(\text{g}) + 3\text{H}_2\text{O} \,(\text{g}); \quad (K)\)
Using the given equilibrium constants, \(K_1, \ K_2 \,\ and \ K_3\), we can write the equilibrium constant K as:
\(K = K_2 \times \frac{1}{K_1} \times (K_3)^3\)
\(K = \frac{K_2 \times (K_3)^3}{K_1}\)
So, the correct option is (A): \(\frac{K_2 \times (K_3)^3}{K_1}\)
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 KEAM exam
- A Spherical ball is subjected to a pressure of 100 atmosphere. If the bulk modulus of the ball is 1011 N/m2,then change in the volume is:
- KEAM - 2023
- mechanical properties of solids
- Two identical solid spheres,each of radius 10cm,are kept in contact. If the mass of inertia of this system about the tangent passing through the point of contact is 0. Then mass of each sphere is:
- KEAM - 2023
- System of Particles & Rotational Motion
- The value of a(≠0) for which the equation \(\frac{1}{2}(x-2)^2+1=\sin(\frac{a}{x})\) holds is/are
- KEAM - 2023
- Trigonometric Functions
- \(∫xlog(1+x^2)dx=\)?
- KEAM - 2023
- Integration by Parts
- A uniform thin rod of mass 3kg has a length of 1m. If a point mass of 1kg is attached to it at a distance of 40cm from its center,the center of mass shifts by a distance of:
- KEAM - 2023
- System of Particles & Rotational Motion
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.