Two solids dissociate as follows
$ {A(s) <=> B(g) + C(g) ; K_{p_1} = x \; atm^2}$
$ {D(s) <=> C(g) + E(g) ; K_{p_2} = y \; atm^2}$
The total pressure when both the solids dissociate simultaneously is :
- $x^2 + y^2 \; atm$
- $ 2 x^2 + y^2 \; atm$
- $2 (\sqrt{x + y} ) atm$
- $\sqrt{x + y} \,atm$
The Correct Option is C
Solution and Explanation
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
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Top JEE Main Law Of Chemical Equilibrium And Equilibrium Constant Questions
- $\text{The following concentrations were observed at 500 K for the formation of NH}_3 \text{ from N}_2 \text{ and H}_2\text{. At equilibrium: [N}_2] = 2 \times 10^{-2} \, \text{M, [H}_2] = 3 \times 10^{-2} \, \text{M, and [NH}_3] = 1.5 \times 10^{-2} \, \text{M. Equilibrium constant for the reaction is \_\_\_\_\_\_.}$
- The following reaction occurs in the Blast furnace where iron ore is reduced to iron metal:
\(\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightleftharpoons 2\text{Fe}(l) + 3\text{CO}_2(g)\)
Using Le Chatelier's principle, predict which one of the following will not disturb the equilibrium. - The equilibrium $\mathrm{Cr_2O_7^{2-}} \rightleftharpoons 2\mathrm{CrO_4^{2-}}$ is shifted to the right in:
- For the given hypothetical reactions, the equilibrium constants are as follows:
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The equilibrium constant for the reaction
\[ X \rightleftharpoons W\]
is: 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: \(R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1}\)
Top JEE Main Questions
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What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
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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.