Question

The equilibrium constant Kp for the reaction I2(g) + Br2(g) ⇀↽ 2 IBr(g) + 11.7 kJ is 280 at 150◦C. Suppose that a quantity of IBr is placed in a closed reaction vessel and the system is allowed to come to equilibrium at 150◦C. When equilibrium is established, the pressure of IBr is 0.200 atm. What is the pressure of I2 at equilibrium?

Answer 1

Answer:

$\large \boxed{\text{0.0120 atm }}$

Explanation:

The balanced equation is

I₂(g) + Br₂(g) ⇌ 2IBr(g)

Data:

      Kp = 280

p(IBr) = 0.200 atm

1. Set up an ICE table.

Let p = the initial pressure of IBr. Then

$\begin{array}{ccccccc}\rm \text{I}_{2}& + & \text{Br}_{2} & \, \rightleftharpoons \, & \text{2IBr} & & \\0 & & 0 & & p & & \\+x & & +x & & -2x & &\\x & & x} & & 280 & & \\\end{array}$

2. Calculate p(I₂)

$\begin{array}{rcl}K_{\text{p}}&=&\dfrac{p_{\text{IBr}}^{2}} {p_{\text{I}_2}^{2}}\\\\280&=&{\dfrac{0.200^{2}}{x^{2}}&&\\\\280x^{2} & = &0.0400\\x^{2} & = &\dfrac{0.0400}{280 }\\\\& = & 1.429 \times 10^{-4}\\x & = & \textbf{0.0120 atm}\\\end{array}\\\text{The partial pressure of iodine is \large \boxed{\textbf{0.0120 atm }} }$}

Check:

$\begin{array}{rcl}{\dfrac{0.200^{2}}{0.0120^{2}}}&=&280\\\\280& =& 280\\\end{array}$