Question

For the equilibrium

Answer 1

Answer:

$\large \boxed{\text{0.091 atm }}$

Explanation:

The balanced equation is

I₂(g) + Br₂(g) ⇌ 2IBr(g)

Data:

   Kc = 8.50 × 10⁻³

n(IBr) = 0.0600 mol

     V = 1.0 L

1. Calculate [IBr]

$\text{[IBr]} = \dfrac{\text{0.0600 mol}}{\text{1.0 L}} = \text{0.0600 mol/L}$

2. Set up an ICE table.

$\begin{array}{ccccccc}\rm \text{I}_{2}& + & \text{Br}_{2} & \, \rightleftharpoons \, & \text{2IBr} & & \\0 & & 0 & &0.0600 & & \\+x & & +x & &- 2x & & \\x & & x & & 0.0600 - 2x & & \\\end{array}$

3. Calculate [I₂]

$\begin{array}{rcl}K_{\text{c}}&=&\dfrac{\text{[IBr]}^{2}} {\text{[I _{2} ][Br] _{2} }}\\\\8.50 \times 10^{-2}&=&{\dfrac{(0.0600 - 2x)^{2}}{x^{2}}}& &\\\\0.2915x & = &{\dfrac{0.0600 - 2x}{x}}& &\\\\0.2915x & = &0.0600 - 2x\\\\2.2915x & = & 0.0600\\x & = & \textbf{0.026 18 mol/L}\\\end{array}$

4. Convert the temperature to kelvins

T = (150 + 273.15) K = 423.15 K

5. Calculate p(I₂)

$\begin{array}{rcl}\\pV & = & nRT\\p & = & cRT\\p & = & \text{0.026 18 mol} \cdot \text{L}^{-1}\times \text{0.082 06 L} \cdot \text{atm} \cdot \text{K}^{-1} \text{mol}^{-1} \times \text{423.15 K}\\& = & \textbf{0.91 atm}\\\end{array}\\\text{The partial pressure of iodine is \large \boxed{\textbf{0.91 atm}} }$