Hydrogen gas and gaseous iodine will react to form hydrogen iodide, as described by the following chemical equation. H2(g)+I2(g)↽−−⇀HI(g)Kc(400∘C)=50.0A ssume that all of the HI(g) is removed from a vessel containing this reaction, and equilibrium is re-established. What will be the new equilibrium concentration of HI if the equilibrium concentrations of H2 and I2 are both equal to 0.450M?

3.18 M

Explanation:

Given data

[H₂]eq = [I₂]eq = 0.450 M

[HI]eq = ?

Let's consider the following reaction at equilibrium.

H₂(g) + I₂(g) ⇄ 2 HI(g)      Kc(400°C) = 50.0

The concentration equilibrium constant (Kc) is:

Kc = [HI]²eq / [H₂]eq × [I₂]eq

[HI]²eq = Kc × [H₂]eq × [I₂]eq

[HI]eq = √(Kc × [H₂]eq × [I₂]eq)

[HI]eq = √(50.0 × 0.450 × 0.450)

[HI]eq = 3.18 M

hope this

explanation:

covalent bonds

consist of pairs of electrons shared by two atoms, and bind the atoms to fixed orientation and requre high energies to break them

ionic bonds

when two atoms differ considerably in their electronegativity, then one of the atoms will lose its electron to the other atom. this results in a positively charged ion(cation) and a negatively charged ion (anion)