Answer to Question #177963 in General Chemistry for Vince

Question

3. Determine the value of the equilibrium constant, Q for the chemical reaction given below, if 0.108 moles of H2, 0.087 of Iodine, 0.020 of Ni , Si with 0.110 and water vapor are in a 6 liters drum and at an equilibrium.

H2= 0.108

I= 0.087

Ni= 0.020

Si= 0.110

6 liters

substitute the value of the compounds

Answer 1

volume of solution = 6L

moles of H₂ = 0.108 moles

Concentration of H₂ = 0.108/6 = 0.018

moles of iodine I₂ = 0.087 moles

Concentration of I₂ = 0.087/6 = 0.0145

moles of Ni = 0.020 moles

Concentration of Ni = 0.020/6 = 0.0033

moles of Si = 0.010 moles

Concentration of Si = 0.010/6 = 0.0016

Now equilibrium constant = $\frac{[H_2][I_2]}{[Ni][Si]}$

= $\frac{[0.018][0.145]}{[0.0033][0.0016]}$ = 49.43