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 Questions on Law Of Chemical Equilibrium And Equilibrium Constant
- Consider the equilibrium: \[ \text{CO(g)} + \text{3H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)} \] If the pressure applied over the system increases by two fold at constant temperature then:
- JEE Main - 2025
- Chemistry
- Law Of Chemical Equilibrium And Equilibrium Constant
- For the reaction: \[ A_2(g) \rightleftharpoons B_2(g) \] The equilibrium constant \( K_c \) is given as 99.0. In a 1 L closed flask, two moles of \( B_2(g) \) is heated to T(K). What is the concentration of \( B_2(g) \) (in mol L\(^{-1}\)) at equilibrium?
- AP EAPCET - 2024
- Chemistry
- Law Of Chemical Equilibrium And Equilibrium Constant
- At equilibrium for the reaction \[ A_2 (g) + B_2 (g) \rightleftharpoons 2AB (g) \] The concentrations of \( A_2 \), \( B_2 \), and \( AB \) respectively are \( 1.5 \times 10^{-3}M \), \( 2.1 \times 10^{-3}M \), and \( 1.4 \times 10^{-3}M \). What will be \( K_p \) for the decomposition of AB at the same temperature?
- AP EAMCET - 2024
- Chemistry
- Law Of Chemical Equilibrium And Equilibrium Constant
- 10 g of a metal (M) reacts with oxygen to form 11.6 g of oxide. What is the equivalent weight of \( M \)?
- AP EAMCET - 2024
- Chemistry
- Law Of Chemical Equilibrium And Equilibrium Constant
- The equilibrium constant for the dissociation of HI at 773 K is:
- AP EAPCET - 2024
- Chemistry
- Law Of Chemical Equilibrium And Equilibrium Constant
Questions Asked in JEE Main exam
- The sum of the squares of all the roots of the equation \( x^2 + [2x - 3] - 4 = 0 \) is:
- JEE Main - 2025
- Quadratic Equations
- Let ABCD be a trapezium whose vertices lie on the parabola \( y^2 = 4x \). Let the sides AD and BC of the trapezium be parallel to the y-axis. If the diagonal AC is of length \( \frac{25}{4} \) and it passes through the point \( (1, 0) \), then the area of ABCD is:
- JEE Main - 2025
- Coordinate Geometry
- A molecule (P) on treatment with acid undergoes rearrangement and gives (Q). (Q) on ozonolysis followed by reflux under alkaline condition gives (R). The structure of (R) is given below. The structure of (P) is:
- JEE Main - 2025
- Organic Chemistry
- The product B formed in the following reaction sequence is: \[ {C}_6{H}_5{CN} \xrightarrow{{HCl}} (A) \xrightarrow{{AgCN}} (B) \]
- JEE Main - 2025
- Organic Chemistry
- Let integers \( a, b \in [-3, 3] \) be such that \( a + b \neq 0 \). \(\text{Then the number of all possible ordered pairs}\) \( (a, b) \), \(\text{for which}\) \[ \left| \frac{z - a}{z + b} \right| = 1 \quad \text{and} \quad \left| \begin{matrix} z + 1 & \omega & \omega^2 \\ \omega^2 & 1 & z + \omega \\ \omega^2 & 1 & z + \omega \end{matrix} \right| = 1, \] \(\text{is equal to:}\)
- JEE Main - 2025
- complex numbers
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.