Question

$15$ moles of $H_2$ and $5.2$ moles of $I_2$ are mixed and allowed to attain equilibrium at $500^\circ$C. At equilibrium, the concentration of $HI$ is found to be $10$ moles. The equilibrium constant for the formation of $HI$ is

• A. 50
• B. 15
• C. 100
• D. 25

Answer 1

Answer: (A)

50

Answer 2

The equilibrium constant (HI)2/ [h2] [I2]= 1010/100.2 = 50.

Let the equation be: wW + xX ⇔ yY + zZ

The equilibrium constant in a reaction can be calculated using the following formula:

Kc = [Y]Y [Z]Z / [W]W [X]X

Equilibrium Constant for a Homogeneous Reaction N2 (g) + 3 H2 (g) ⇔ 2NH3 (g)