Question

The following reaction was performed in a sealed vessel at 734 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only H2 and I2 were present at concentrations of [H2]=3.75M and [I2]=2.15M . The equilibrium concentration of I2 is 0.0800 M . What is the equilibrium constant, Kc, for the reaction at this temperature? Express your answer numerically.

Answer 1

Answer:

128

Explanation:

Step 1: Write the balanced equation

H₂(g) + I₂(g) ⇌ 2 HI(g)

Step 2: Make an ICE chart

         H₂(g) + I₂(g) ⇌ 2 HI(g)

I         3.75     2.15         0

C         -x         -x         +2x

E      3.75-x   2.15-x      2x

Step 3: Find the value of x

Since the concentration of I₂ at equilibrium is 0.0800 M,

2.15M-x = 0.0800 M

x = 2.07 M

Step 4: Calculate the concentrations at equilibrium

[H₂] = 3.75-x = 3.75-2.07 = 1.68 M

[I₂] = 0.0800 M

[HI] = 2x = 2(2.07) = 4.14 M

Step 5: Calculate the equilibrium constant Kc

$Kc= \frac{[HI]^{2} }{[H_2]\times [I_2]} = \frac{4.14^{2} }{1.68\times 0.0800} = 128$