What is the minimum volume of water required to dissolve 1g of calcium sulfate at 298 K? (For calcium sulfate, Ksp is 9.1 × 10–6).
Solution and Explanation
CaSO4(s) ↔ Ca2+(aq) + SO2-4(aq)
Ksp = [Ca2+][SO2-4]
Let the solubility of CaSO4 be s.
Then, Ksp = s2 = 9.1 × 10-6 = s2
s = 3.02 × 10-3 mol / L
Molecular mass of CaSO4 = 136 g/mol
Solubility of CaSO4 in gram/L = 3.02 × 10-3 × 136 = 0.41 g/L
This means that we need 1L of water to dissolve 0.41g of CaSO4 Therefore, to dissolve 1g of CaSO4 we require = \(\frac{1}{0.41}\) L = 2.44L of water.
Top Questions on Law Of Chemical Equilibrium And Equilibrium Constant
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]
- JEE Main - 2025
- Chemistry
- 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
- At 700 K, the Equilibrium constant value for the formation of HI from $ H_2 $ and $ I_2 $ is 49.0. 0.7 mole of HI(g) is present at equilibrium. What will be the concentrations of $ H_2 $ and $ I_2 $ gases if we initially started with HI(g) and allowed the reaction to reach equilibrium at the same temperature?
- COMEDK UGET - 2024
- Chemistry
- 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
- Chemistry
- Law Of Chemical Equilibrium And Equilibrium Constant
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
- Chemistry
- Law Of Chemical Equilibrium And Equilibrium Constant
Questions Asked in CBSE Class XI exam
- Describe the structure of the following with the help of labelled diagrams.
(i) Nucleus
(ii) Centrosome- CBSE Class XI
- An Overview Of Cell
- Convert the following into basic units:
- 28.7 pm
- 15.15 pm
- 25365 mg
- CBSE Class XI
- Percentage Composition
- Using s, p, d notations, describe the orbital with the following quantum numbers.
- n=1, l=0
- n = 3, l=1
- n = 4, l =2
- n=4, l=3
- CBSE Class XI
- Developments Leading to the Bohr’s Model of Atom
- Three reasons why the author’s grandmother was disturbed when he started going to the city school.
- CBSE Class XI
- The Portrait of a lady
- The author’s grandmother was a religious person. What are the different ways in which we come to know this?
- CBSE Class XI
- The Portrait of a lady
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.