Consider the following hypothetical reaction.
$A+B \rightleftharpoons C+D ; \,\,\,K_{c}=1 \times 10^{-4}$
Which of the following statements is true for this reaction ? In this reaction,
- products predominate
- reactants predominate
- equilibrium exists
- nothing can be said
The Correct Option is B
Solution and Explanation
Top Questions on Law Of Chemical Equilibrium And Equilibrium Constant
- For the reaction in equilibrium:
N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = -Q
Reaction is favoured in forward direction by:- NEET (UG) - 2024
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- At a given temperature and pressure, the equilibrium constant values for the equilibria are given below:
\[ 3A_2 + B_2 \rightleftharpoons 2A_3B, \, K_1 \]
\[ A_3B \rightleftharpoons \frac{3}{2}A_2 + \frac{1}{2}B_2, \, K_2 \]
The relation between \( K_1 \) and \( K_2 \) is:- NEET (UG) - 2024
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- At \(-20^\circ \text{C}\) and 1 atm pressure, a cylinder is filled with an equal number of \(H_2\), \(I_2\), and \(HI\) molecules for the reaction:
\[H_2(g) + I_2(g) \rightleftharpoons 2HI(g)\] The \(K_P\) for the process is \(x \times 10^{-1}\).
\(x = ___________)
Given: \(R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1}\)- JEE Main - 2024
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- For the given hypothetical reactions, the equilibrium constants are as follows:
\[ X \rightleftharpoons Y; \, K_1 = 1.0 \]
\[ Y \rightleftharpoons Z; \, K_2 = 2.0 \]
\[ Z \rightleftharpoons W; \, K_3 = 4.0 \]
The equilibrium constant for the reaction
\[ X \rightleftharpoons W\]
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Concepts Used:
Law of Chemical Equilibrium
Law of Chemical Equilibrium states that at a constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants each raised to a power equal to the corresponding stoichiometric coefficients as represented by the balanced chemical equation.
Let us consider a general reversible reaction;
A+B ↔ C+D
After some time, there is a reduction in reactants A and B and an accumulation of the products C and D. As a result, the rate of the forward reaction decreases and that of backward reaction increases.
Eventually, the two reactions occur at the same rate and a state of equilibrium is attained.
By applying the Law of Mass Action;
The rate of forward reaction;
Rf = Kf [A]a [B]b
The rate of backward reaction;
Rb = Kb [C]c [D]d
Where,
[A], [B], [C] and [D] are the concentrations of A, B, C and D at equilibrium respectively.
a, b, c, and d are the stoichiometric coefficients of A, B, C and D respectively.
Kf and Kb are the rate constants of forward and backward reactions.
However, at equilibrium,
Rate of forward reaction = Rate of backward reaction.
Kc is called the equilibrium constant expressed in terms of molar concentrations.
The above equation is known as the equation of Law of Chemical Equilibrium.