Question

A 2.00L flask was filled with 4.00 mol of HI at a certain temperature and given sufficient time to react. At equilibrium the concentration of H2 was 0.400 M. Find the equilibrium concentrations of I2 and HI and then find the Keq at this temperature.

Answer 1

Answer:

The equilibrium concentration of I₂ is 0.400 M  and HI is 1.20 M, the Keq will be 0.112.

Explanation:

Based on the given information, the equilibrium reaction will be,  

2HI (g) ⇔ H₂ (g) + I₂ (g)

It is given that 4.00 mol of HI was filled in a flask of 2.00 L, thus, the concentration of HI will be,  

= 4.00 mol/2.00 L

= 2.00 mol/L

Based on the reaction, the initial concentration of 2HI is 2.00, H₂ is 0 and I₂ is O. The change in the concentration of 2HI is -x, H₂ is x and I₂ is x. The equilibrium concentration of 2HI will be 0.200-x, H₂ is x and I₂ is x.  

It is given that at equilibrium, the concentration of H₂ or x is 0.400 M.  

Now the equilibrium concentration of HI will be,  

= 2.00 -2x  

= 2.00 - 2 × 0.400

= 1.20 M

The equilibrium concentration of I₂ will be,  

I₂ = x  

= 0.400 M

The equilibrium constant (Keq) will be,  

Keq = [H₂] [I₂] / [HI]²

= (0.400) (0.400) / (1.20)²

= 0.112

Thus, the Keq of the reaction will be 0.112.