Question

Read the reaction and the statement.

H₂(g) + I₂(g) ⇌ 2HI(g)
The concentration of H₂, I₂, and HI are measured as 0.04 M, 0.08 M, and 0.12 M at equilibrium and 200 K. What is the value of the equilibrium constant?
a) 9.0
b) 2.5
c) 1.0
d) 4.5

Answer 1

The equilibrium concentrations are approximately: [H₂]eq ≈ 0.9374 M, [I₂]eq ≈ 1.9374 M, and [HI]eq ≈ 0.1252 M.

Step-by-step explanation:

In order to find the equilibrium concentrations of H₂, I₂, and HI in moles/L, we need to use the equilibrium constant expression. The given equilibrium constant (Kc) value for the reaction is 50.5. The equilibrium concentrations of H₂, I₂, and HI are not provided, so we need to solve for them using an ICE (Initial Change Equilibrium) table.

1. Let x be the change in concentration for H₂, I₂, and HI.
2. The initial concentration of H₂ is 1.00 mole/L, but since it reacts with I₂, its concentration will decrease by x.
3. The initial concentration of I₂ is 2.00 moles/L, and it reacts with H₂ to form HI, so its concentration will also decrease by x.
4. The initial concentration of HI is 0 M, so its concentration will increase by 2x.

Using the equilibrium constant expression, we can write the equation: Kc = (([HI]eq)²)/(([H₂]eq)([I₂]eq)). Plugging in the given values and using the information from the ICE table, we can solve for the equilibrium concentrations of H₂, I₂, and HI.

H₂: [H₂]eq = 1.00 - x
I₂: [I₂]eq = 2.00 - x
HI: [HI]eq = 2x
Kc = ((0.75 M)²) / ((0.20 M)(x))

50.5 = 0.75² / (0.20)(x)
50.5 = 0.5625 / (0.20)(x)
50.5 = 2.8125 / (x)
x = 0.0626

Therefore, the equilibrium concentrations of H₂, I₂, and HI in moles/L are approximately:

H₂: [H₂]eq ≈ 1.00 - x ≈ 1.00 - 0.0626 ≈ 0.9374 M
I₂: [I₂]eq ≈ 2.00 - x ≈ 2.00 - 0.0626 ≈ 1.9374 M
HI: [HI]eq ≈ 2x ≈ 2(0.0626) ≈ 0.1252 M