For the following gas phase equilibrium reaction at constant temperature, NH3(g) 2 N2(g) + 3 2 H2(g) If the total pressure is 3 atm and the pressure equilibrium constant (KP) is 9 atm, then the degree

Question

For the following gas phase equilibrium reaction at constant temperature, NH3(g) 2 N2(g) + 3 2 H2(g) If the total pressure is 3 atm and the pressure equilibrium constant (KP) is 9 atm, then the degree of dissociation is given as (x × 10 –2) –1/2. The value of x is ______ (Nearest integer)

Prashant Kotian · Accepted Answer

Correct answer: .. 3(g) 2(g) 2(g) 3 NH N H 2  t = 0 1mole – – t= teq 1-  1/2 3/2 T P P 2 k (1 )                        T P 3atm      1/2 3/2 1/2 (3) 2 (1 )                   2 9 2 1          2 4   5 4    = 0.8  = (0.8) 1/2  = 1/2 0.8         = [125 × 10 –2] –1/2 x = 125.

💡 Solution & Explanation

Practice 22,000+ questions like this