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Equilibrium constant
Appears in
Concepts tested here
- Calculating Kc
- Combining equilibria
- Equilibrium concentrations
- K for a scaled equation
- K for reversed and halved reaction
- Magnitude of K
All Questions
2015 AIPMT-I 1 question
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If the value of an equilibrium constant for a particular reaction is $1.6\times 10^{12}$, then at equilibrium the system will containFor $A + B \rightleftharpoons C + D$, $K_c = \frac{[C][D]}{[A][B]}$
A very high $K_c$ means the numerator (product concentrations) is large, so the system contains mostly products.
2015 AIPMT-II 1 question
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If the equilibrium constant for $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$ is K, the equilibrium constant for $\frac{1}{2}N_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons NO(g)$ will be:$N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$; K
$\frac{1}{2}N_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons NO(g)$; K'
When a reaction is multiplied by $\frac{1}{2}$, the equilibrium constant is raised to the power $\frac{1}{2}$: $K' = K^{1/2}$
2012 AIPMT-MAINS 2 questions
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Given that the equilibrium constant for the reaction $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature? $SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2}O_2(g)$$2SO_2 + O_2 \rightleftharpoons 2SO_3$; $K = 278$
Reversing the reaction gives $\frac{1}{K}$; halving it gives the square root.
$SO_3 \rightleftharpoons SO_2 + \frac{1}{2}O_2$; $K' = \sqrt{\frac{1}{K}} = \sqrt{\frac{1}{278}}$
$K' = 5.99\times10^{-2} \approx 6.0\times10^{-2}$ -
Given the reaction between 2 gases represented by $A_2$ and $B_2$ to give the compound AB(g): $A_2(g) + B_2(g) \rightleftharpoons 2AB(g)$. At equilibrium, the concentration of $A_2 = 3.0\times10^{-3}$ M, of $B_2 = 4.2\times10^{-3}$ M, of AB $= 2.8\times10^{-3}$ M. If the reaction takes place in a sealed vessel at $527^\circ C$, then the value of $K_c$ will be:$A_2 + B_2 \rightleftharpoons 2AB$
$K_c = \frac{[AB]^2}{[A_2][B_2]}$
$K_c = \frac{(2.8\times10^{-3})^2}{(3.0\times10^{-3})(4.2\times10^{-3})} = \frac{7.84}{12.6}$
$K_c = 0.62$
2011 AIPMT-PRE 1 question
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For the reaction $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$, the equilibrium constant is $K_1$. The equilibrium constant is $K_2$ for the reaction $2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)$. What is K for the reaction $NO_2(g) \rightleftharpoons \frac{1}{2}N_2(g) + O_2(g)$?(i) $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$; $K_1$
(ii) $2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)$; $K_2$
Adding (i) and (ii): $N_2(g) + 2O_2(g) \rightleftharpoons 2NO_2(g)$; when reactions are added their constants multiply, so $K = K_1K_2$
Reversing this reaction: $2NO_2(g) \rightleftharpoons N_2(g) + 2O_2(g)$; the constant becomes the reciprocal, $\frac{1}{K_1K_2}$
Dividing the equation by 2: $NO_2(g) \rightleftharpoons \frac{1}{2}N_2(g) + O_2(g)$; the constant is raised to the power $\frac{1}{2}$
$K = \left[\frac{1}{K_1K_2}\right]^{1/2}$
2010 AIPMT-MAINS 1 question
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The reaction $2A(g) + B(g) \rightleftharpoons 3C(g) + D(g)$ is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression$2A + B \rightleftharpoons 3C + D$; initial concentrations: [A] = 1.00, [B] = 1.00, [C] = 0, [D] = 0
At equilibrium [D] = 0.25 M, so the extent of reaction is x = 0.25 M.
[C] = 3x = 0.75 M
[A] = 1.00 − 2x = 1.00 − 0.50 = 0.50 M
[B] = 1.00 − x = 1.00 − 0.25 = 0.75 M
$K = \frac{[C]^3[D]}{[A]^2[B]} = \frac{(0.75)^3(0.25)}{(0.50)^2(0.75)}$