At a certain temperature the rate of this reaction is first order in hi with a rate constant of : 7.21s-1
2hi(g) â h2(g) + i2(g)

suppose a vessel contains hi at a concentration of 0.440m
calculate the concentration of hi in the vessel 0.210 seconds later. you may assume no other reaction is important. round your answer to 2 significant digits.

C HI = 0.1 M 0.210 second later

Explanation:

∴ order (∝) = 1

∴ κ = 7.21 s-1

∴ Cao = 0.440 M initial concentration of HI

∴ t = 0.210 s

rate of this reaction (r):

⇒ - ra = κ.(Ca)∧∝ ∝ = 1

⇒ - ra = κ.Ca = - δCa/δt

⇒ - δCa/Ca = κ.δt

⇒ - ∫ δCa/Ca = κ ∫δt

⇒ Ln Cao/Ca = κ.t

⇒ Cao/Ca = e∧(κ.t)

⇒ Ca = Cao/e∧(κ.t)

⇒ Ca = ( 0.44 ) / e∧(7.21×0,21)

⇒ Ca = 0.0968 M

⇒ Ca = 0.1 M

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."