Answer:
[H2] = 0.0692 M
[I2] = 0.182 M
[HI] = 0.826 M
Explanation:
Step 1: Data given
Kc = 54.3 at 430 °C
Number of moles hydrogen = 0.714 moles
Number of moles iodine = 0.984 moles
Number of moles HI = 0.886 moles
Volume = 2.40 L
Step 2: The balanced equation
H2 + I2 → 2HI
Step 3: Calculate Q
If we know Q, we know in what direction the reaction will go
Q = [HI]² / [I2][H2]
Q= [n(HI) / V]² /[n(H2)/V][n(I2)/V]
Q =(n(HI)²) /(nH2 *nI2)
Q = 0.886²/(0.714*0.984)
Q =1.117
Q<Kc This means the reaction goes to the right (side of products)
Step 2: Calculate moles at equilibrium
For 1 mol H2 we need 1 mol I2 to produce 2 moles of HI
Moles H2 = 0.714 - X
Moles I2 = 0.984 -X
Moles HI = 0.886 + 2X
Step 3: Define Kc
Kc = [HI]² / [I2][H2]
Kc = [n(HI) / V]² /[n(H2)/V][n(I2)/V]
Kc =(n(HI)²) /(nH2 *nI2)
KC = 54.3 = (0.886+2X)² /((0.714 - X)*(0.984 -X))
X = 0.548
Step 4: Calculate concentrations at the equilibrium
[H2] = (0.714-0.548) / 2.40 = 0.0692 M
[I2] = (0.984 - 0.548) / 2.40 = 0.182 M
[HI] = (0.886+2*0.548) /2.40 = 0.826 M
