The gas phase decomposition of hydrogen iodide at 700 K HI(g)½ H2(g) + ½ I2(g) is second order in HI with a rate constant of 1.20×10-3 M-1 s-1. If the initial concentration of HI is 1.09 M, the concentration of HI will be M after 2.36×103 seconds have passed.

C HI = 0.2667 M

Explanation:

decomposition rate HI(g):

∴ a: HI(g)

∴ K = 1.20 E-3 M/srate constant

∴ α = 2second order

∴ Cao = 1.09 Minitial concentration of HI(g)

⇒ Ca = ? t = 2.36 E3 s

⇒ - ra = δCa/δt = K Ca²

⇒ - ∫δCa/Ca² = K∫δt

⇒ [ 1/Ca - 1/Cao ] = K*t

⇒ 1/Ca = (K*t) + 1/Cao

⇒ 1/Ca = ((1.20 E-3/M.s)(2.36 E3 s)) + (1/1.09 M)

⇒ 1/Ca = 2.832/M + 0.9174/M

⇒ 1/Ca = 3.749/M

⇒ Ca = 0.2667 M

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole