Question

In: Chemistry

Consider the following reaction and its equilibrium constant: I2(g) + Br2(g) 2 IBr(g) Kc = 1.1...

Consider the following reaction and its equilibrium constant:

I2(g) + Br2(g) 2 IBr(g) Kc = 1.1 × 102

This reaction mixture contains initially 0.41 M I2 and 0.27 M Br2. Calculate the equilibrium concentration of I2, Br2, and IBr?

Please help with this, Can I find a example similar to this anywhere. I can't figure out some of the parts of this problem.

Expert Solution

Consider the following reaction and its equilibrium constant:

I2(g) + Br2(g) 2 IBr(g) Kc = 1.1 × 102

This reaction mixture contains initially 0.41 M I2 and 0.27 M Br2. Calculate the equilibrium concentration of I2, Br2, and IBr?

Please help with this, Can I find a example similar to this anywhere. I can't figure out some of the parts of this problem.

Ans:

I₂(g)+ Br₂(g) ↔ 2IBr (g) Kc = 1.1 × 10² =110

Kc = [products]/[reactants] = ([IBr]²)/(([I₂])*([Br₂]))

ICE table :

Species...Initial...Change...Equilib.....
I₂...........0.41M...    -x...       0.41-x....
Br₂.........0.27M...   -x...       0.27-x....
IBr......        0.....     +2x...      2x ...

Now just plug in the numbers and calculate Kc:

Kc = [products]/[reactants] = ([IBr]²)/(([I₂])*([Br₂]²))
Kc = ((2x)²)/(( 0.41-x)*( 0.27-x))
110 = ((2x)²)/(( 0.41-x)*( 0.27-x))

110 (0.1107-0.41x-0.27x+x²) = 4x²

110 (0.1107-0.68x+x²) = 4x²

12.177-74.8x+110x² = 4x²

106x²-74.8x+12.177 = 0

If we solve this quadratic equation we will get x=0.254 or x=0.45

x=0.45 is not possible because initial concentration of I₂ is 0.41 M and Br₂ is 0.27 M

Equilibrium concentration of I₂ = 0.41-x = 0.41-0.254 = 0.156 M

Equilibrium concentration of Br₂= 0.27-0.254 = 0.016 M

Equilibrium concentration of IBr = 2*0.254 = 0.508M

queen_honey_blossom answered 2 years ago

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