Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
NTA Abhyas JEE Main2020ChemistryChemical EquilibriumPractice

For a reaction A → B, E a = 10 kJ/mol, ΔH = 5 kJ/mol . Thus potential energy profile for this reaction is

Correct answer

1

Step-by-step solution

Δ H = ( E a ) ƒ – ( E a ) b 5 = 10 – ( E a ) b (E a ) b = 5  kJ/mol . ( E a ) ƒ > ( E a ) b hence option B is correct

Practice Chemical Equilibrium on Quantrex Academy →

More from Chemical Equilibrium

Consider the following reactions in which all the reactants and products are present in gaseous state 2xy x₂ + y₂ K₁ = 2.5 10^5 xy + 1 2 z₂ xyz K₂ = 5 10⁻³ The value of K₃ for the 2026Solid carbon, CaO and CaCO₃ are mixed and allowed to attain equilibrium at T K. CaCO ₃(s) CaO (s) + CO ₂(g) K_ p₁ = 0.08 atm C (s) + CO ₂(g) 2 CO (g) K_ p₂ = 2 atm The partial pres 2026In a closed flask at 600 K, one mole of X₂ Y₄ (g) attains equilibrium as given below : X ₂ Y ₄(g) 2 XY ₂(g) At equilibrium, 75 % X₂ Y₄ (g) was dissociated and the total pressure is 2026One mole each of He and A(g) are taken in a 10 L closed flask and heated to 400 K to establish the following equilibrium. A(g) B(g) . K_c for this reaction at 400 K is 4.0 . The pa 2026The reaction A(g) B(g) + C(g) was initiated with the amount ' a ' of A(g) . At equilibrium it is found that the amount of A(g) remaining is (a - x) at a total pressure of p . The e 2026The values of pressure equilibrium constant recorded at different temperatures for the following equilibrium reaction have been given below A(g) B(g) + C(g) 1 T ( K ⁻¹) ₁₀ K_p 0.05 2026For the following reaction at 50° C and at 2 atm pressure, 2N₂O₅(g) 2N₂O₄(g)+O₂(g) N₂O₅ is 50 % dissociated. The magnitude of standard free energy change at this temperature is x . 2026At T(K) , the equilibrium constant of A₂(g) + B₂(g) C(g) is 2.7 10⁻⁵ . What is the equilibrium constant for 1 3 A₂(g) + 1 3 B₂(g) 1 3 C(g) at the same temperature? 2026 Full Chemical Equilibrium list All NTA Abhyas JEE Main PYQs