# When 2 moles of hydrogen is heated with 2 moles of iodine,2.96 moles of hydrogen iodide is formed.what is the equilibrium constant for the formation of hydrogen iodide?

Jan 19, 2017

$\text{K"_"c} = 4$

#### Explanation:

In this question, we aren't given the equilibrium concentrations of our reagents and products, we have to work it out ourselves using the ICE method. First, we must write out the balanced equation.

$\textcolor{w h i t e}{a a a a a a a a a a a a a a a} \text{H"_2 color(white)(aa)+ color(white)(aa)"I"_2 color(white)(aa)rightleftharpoons color(white)(aa)2"HI}$

Initial moles: $\textcolor{w h i t e}{a a a a a z} 2 \textcolor{w h i t e}{a a a a a a a} 2 \textcolor{w h i t e}{a a a a a a a a a} 0$

Change in moles: $- 1.48 \textcolor{w h i t e}{a a} - 1.48 \textcolor{w h i t e}{a a a} + 2.96$

Equilibrium moles: $\textcolor{w h i t e}{a} 0.53 \textcolor{w h i t e}{z a c a a} 0.53 \textcolor{w h i t e}{a a a a a} 2.96$

So we know how many moles we have of each at equilibrium. Since we aren't given the concentration, we must assume that the number of mols equals the molarity.

"K"_"c" = ["HI"]^2/(["H"_2]["I"_2])=2.96^2/((1.48)(1.48))=4

The equilibrium constant has no units as they cancel out:

cancel("mol"^2)/(cancel(("mol")("mol")))