2.00 mol of H2(g) and 1.00 mol of I2(g) are placed in a 1.00 L container, and they react to form HI(g). At equilibrium, it is found that 1.80 moles of HI(g) are present in the container. Calculate K for the reaction: H2(g) + I2(g) ⇄ 2 HI(g)

The value of equilibrium constant is 29.45.

Explanation:

Moles of hydrogen gas = 2.00 mol

Concentration of  hydrogen gas =

Moles of iodine gas = 1.00 mol

Concentration of  iodine gas =

initially

2.00 M   1.00 M    1.00 M

At equilibrium:

(2.00-x/2)    (1.00-x/2)      x

Moles of  HI at equilibrium = 1.80 M

Concentration of HI at equilibrium =

The expression of an equilibrium constant is given by ;

Putting x equal to 1.80 M.

The value of equilibrium constant is 29.45.

answer: option (b) is the correct answer.

explanation:

a covalent bond is formed by sharing of electrons. whereas polarity arises due to difference in electronegativity of two atoms.

therefore, more is the electronegativity of an atom which combines with the titanium, more readily it will induce a partial positive and negative charge.

thus, out of the given options fluorine is most electronegative and has an incomplete octet. therefore, it will readily combine with titanium to share the electrons and complete its octet.