Question

PLS I REALLY NEED HELP DUE FODAY
For the reaction of hydrogen gas with iodine gas to make hydrogen iodide gas, H2 + 12 → 2HI, you have the following concentrations at equilibrium: [HQ] = 4.5647 x 10-3 M, [12] = 7.378 x 10-4 [HI] = 1.3544 x 10-2 M. M, and What is the equilibrium constant?
0.4997
0.54
46.33
54.47​

Answer 1

Answer: The equilibrium constant for the reaction is 54.47

Explanation:

The equilibrium constant is defined as the ratio of the concentration of products to the concentration of reactants raised to the power of the stoichiometric coefficient of each. It is represented by the term $K_{eq}$

The given chemical equation follows:

$H_2+I_2\rightleftharpoons 2HI$

The expression for equilibrium constant will be:

$K_{eq}=\frac{[HI]^2}{[H_2][I_2]}$

We are given:

$[HI]_{eq}=1.3544\times 10^{-2}M$

$[H_2]_{eq}=4.5647\times 10^{-3}M$

$[I_2]_{eq}=7.378\times 10^{-4}M$

Putting values in above expression, we get:

$K_{eq}=\frac{(1.3544\times 10^{-2})^2}{(4.5647\times 10^{-3})(7.378\times 10^{-4})}\\\\K_{eq}=54.47$

Hence, the equilibrium constant for the reaction is 54.47