Answer:
0.628 mol·L⁻¹
Explanation:
The balanced equation is
H₂ + I₂ ⇌ 2I ₂
Data:
Kc = 53.3
[H₂] = 0.400 mol·L⁻¹
[I₂] = 0.400 mol·L⁻¹
1. Set up an ICE table.
[tex]\begin{array}{ccccccc}\rm \text{H}_{2}& + & \text{I}_{2} & \, \rightleftharpoons \, & \text{2HI} & & \\0.400 & & 0.400 & & 0 & & \\-x & & -x & & +2x & & \\0.400 - x & & 0.400 - x & & 2x & & \\\end{array}[/tex]
2. Calculate the equilibrium concentrations
[tex]K_{\text{c}} = \dfrac{\text{[HI]$^{2}$}}{\text{[H$_{2}$][I$_2$]}} = \dfrac{(2x)^{2}}{(0.400 - x)^{2}} = 53.3\\\\\begin{array}{rcl}\dfrac{(2x)^{2}}{(0.400 - x)^{2}} &=& 53.3\\ \dfrac{2x }{0.400 - x} & = & 7.301\\\\2x & = & 7.301(0.400 - x)\\2x & = & 2.920 - 7.301x\\9.301x & = & 2.920\\x & = & \dfrac{2.920}{9.301}\\\\x & = & \mathbf{0.3140}\\\end{array}[/tex]
[HI] = 2x mol·L⁻¹ = 2 × 0.3140 mol·L⁻¹ = 0.628 mol·L⁻¹
The equilibrium constant for the reaction H₂(g) + I₂(g) ⇄ 2HI(g) is 53.3, so when 0.400 moles of H₂ and I₂ are placed in a container of 1.00 L, the concentration of HI present at equilibrium is 0.62 mol/L.
The balanced reaction is:
H₂(g) + I₂(g) ⇄ 2HI(g) (1)
The equilibrium constant for the above reaction is:
[tex] K_{c} = \frac{[HI]^{2}}{[H_{2}][I_{2}]} = 53.3 [/tex] (2)
Before the reaction, the concentration of H₂ and I₂ is:
[tex] [H_{2}] = [I_{2}] = \frac{0.400 mol}{1 L} = 0.400 mol/L [/tex]
At equilibrium we have:
H₂(g) + I₂(g) ⇄ 2HI(g) (3)
0.4-x 0.4-x 2x
Entering the values of equation (3) into equation (2), we have:
[tex] 53.3 = \frac{(2x)^{2}}{(0.400 - x)(0.400 - x)} [/tex]
[tex] 53.3(0.400 - x)^{2} - 4x^{2} = 0 [/tex]
After solving the above equation for x, we get:
[tex] x_{1} = 0.31 mol/L [/tex]
[tex] x_{2} = 0.55 mol/L [/tex]
Since the concentrations of H₂ and I₂ could not be negative, we will take the x₁ value (with the x₂ value, we have [H₂] = [I₂] = 0.400 - 0.55 = -0.15).
Hence, the concentration of HI is:
[tex][HI] = 2x = 2*0.31 mol/L = 0.62 mol/L[/tex]
Therefore, the concentration of HI present at equilibrium is 0.62 mol/L.
Learn more about equilibrium constants here:
I hope it helps you!
