# Find K_c for the reaction below (see details)?

## \sf{2ICl(g)\harrI_2(g)+Cl_2(g) A 0.0682-gram sample of $\text{ICl (g)}$ is placed in a 625-mL reaction vessel at 628 K. When equilibrium is reached between the $\text{ICl (g)}$ and ${I}_{2} \text{ (g)}$ formed by dissociation, 0.0383-grams of ${I}_{2}$ are present. What is ${K}_{c}$ for this reaction? Sorry if this has been asked before!

Aug 5, 2018

$k = 3 \times {10}^{-} 4$

#### Explanation:

The balanced equation of the given reversible gaseous reaction

\sf{2ICl(g)rightleftharpoonsI_2(g)+Cl_2(g)

Molar masses of reactant and products

$I C l \to 162.5 \text{ } g \cdot m o {l}^{-} 1$

${I}_{2} \to 254 \text{ } g \cdot m o {l}^{-} 1$

$C {l}_{2} \to 71 \text{ } g \cdot m o {l}^{-} 1$

Volume of reaction vessel $V = 625 m L = 0.625 L$

ICE Table

$\text{ "" "\sf{2ICl(g)" "" "rightleftharpoons" "I_2(g)" "+" } C {l}_{2} \left(g\right)$

$I \text{ " " "\alpha" "mol" "" "" "" "0" "mol" "" "0" } m o l$

$C \text{ "-2x" "mol" "" "" "x" "mol" "" "x" } m o l$

$E \text{ "\alpha-2x" "mol" "" "" "x" "mol" "" "x" } m o l$

By the problem initial amount of $I C l \left(g\right)$

$\setminus \alpha = \left(0.0682 \text{ g "ICl)/(162.5" g/mol } I C l\right) \approx 4.2 \times {10}^{- 4} m o l$

Amount of ${I}_{2} \left(g\right)$ as well as $C {l}_{2} \left(g\right)$ in equilibrium mixture

$x = \left(0.0383 \text{ g")/(254" g/mol}\right) \approx 1.5 \times {10}^{-} 4 m o l$

The equilibrium constant ${K}_{c}$ of the reaction

K_c=([I_2(g)][Cl_2(g)])/([ICl (g)]

$= \frac{\frac{x}{V} \cdot \frac{x}{V}}{\frac{\setminus \alpha - 2 x}{V}}$

$= {x}^{2} / \left(V \left(\setminus \alpha - 2 x\right)\right)$

$= {\left(1.5 \times {10}^{-} 4\right)}^{2} / \left(0.625 \left(4.2 \times {10}^{- 4} - 2 \cdot 1.5 \times {10}^{-} 4\right)\right)$

$= \frac{2.25 \times {10}^{-} 8}{0.625 \times 1.2 \times {10}^{-} 4} = 3 \times {10}^{-} 4$