Question

hydrogen and lodine react to form hydrogen iodide, like this:

mathrm{h}_{2}(g)+mathrm{i}_{2}(g)
ightarrow 2 mathrm{hi}(g)

also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen, iodine, and hydrogen iodide has the following composition:

calculate the value of the equilibrium constant ( k_{c} ) for this reaction. round your answer to 2 significant digits.

k_{c}=

Explanation:

Step1: Write the equilibrium constant expression

For the reaction \(H_{2}(g)+I_{2}(g)
ightleftharpoons 2HI(g)\), the equilibrium constant \(K_{c}\) is given by \(K_{c}=\frac{[HI]^{2}}{[H_{2}][I_{2}]}\)

Step2: Substitute the equilibrium concentrations

Substitute \([H_{2}]=1.6M\), \([I_{2}]=1.1M\), and \([HI]=1.8M\) into the expression:
\(K_{c}=\frac{(1.8)^{2}}{1.6\times1.1}\)
First, calculate \((1.8)^{2}=3.24\) and \(1.6\times1.1 = 1.76\)
Then \(K_{c}=\frac{3.24}{1.76}\approx1.8\)

Answer:

\(1.8\)