alleryFor the reaction H2(g) + I2(g) ⇌ 2HI(g) with Kc = 50 at 800 K, if [H2] = [I2] = 0.01 M initially, equilibrium [HI]:
A0.020
B0.005
C0.014
D0.010
Answer & Solution
Correct answer: C. 0.014
Let x = HI formed.
Then [H2] = [I2] = 0.01 - x/2. Kc = x²/((0.01-x/2)²) = 50. So x/(0.01-x/2) = √50 ≈ 7.07. x ≈ 7.07(0.01 - x/2) = 0.0707 - 3.535x.
4.535x = 0.0707, x ≈ 0.0156. Approximate [HI] ≈ 0.014 M.
Keep practising Unit 7A1: Equilibrium Constant and Dynamic Equilibrium
Browse all Unit 7A1: Equilibrium Constant and Dynamic Equilibrium questions for JEE Advanced, or explore more Chemistry topics.
Related questions
Degree of dissociation α of weak acid HA (Ka = 1.8 × 10⁻⁵) in 0.01 M solution:In a saturated solution of AgCl, [Ag+] = 1.26 × 10⁻⁵. If KNO3 is added (no common ion), soAt equilibrium, the ratio of forward to reverse reaction rates:At 298 K, water has Kw = 10⁻¹⁴. Concentration of H+ in pure water:If K = 1 for a reaction, the standard free energy change ΔG°:For the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), at equilibrium increasing pressure:For the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g), if initial pressure of PCl5 is P0 and degree Solubility of CaF2 in water (Ksp = 4.0 × 10⁻¹¹):