The rate constant for the following equation is 0.0032 M1 · s 1 . We start with

Question

The rate constant for the following equation is 0.0032 M1 · s 1 . We start with 0.245 mol C2F4 in a 3 liter container, with no C4F8 initially present.

a. 2 C2F4 C4F8; What will be the concentration of C2F4 after 3 hours? Answer in units of M

b. What will be the concentration of C4F8 after 3 hours? Answer in units of M.

c. What is the half-life of the reaction for the initial C2F4 concentration in part 1? Answer in units of s.

d. How long will it take for half of the C2F4 that remains after 3 hours (from part 1) to disappear? Answer in units of s.

Explanation / Answer

Given rate costant, k = 0.0032 M-1s-1

From the units of the rate constant , it is second order reaction.

For second order reaction: 1/[A] = 1/[Ao] + k (delta t) ---- (1)

a) Initial concentration of C2F4 ,Ao = 0.245mol/3L ; t = 3 x 3600 s ; k = 0.0032 M-1 s-1

    1/[A] = 1/ [Ao] + k (delta t)

substituting the above values we get

    1/[A] = 1/46.8 M

Concentration of C2F4 after 3 hrs , [A] = 0.02136 M

b) amount of C2F4 remaining after 3 hrs = 0.02136 x 3 mol = 0.06408 mol

amount of the C2F4 converted = 0.245 - 0.06408 = 0.1809 mol

for 2 mol C2F4 1 mol C4F8 is formed , for 0.1809 mol C2F4 amount of C4F8 formed = 0.1809/2 = 0.09045 mol

Amount of C4F8 formed in M = 0.09045/3 = 0.0315 M

c) at half life , [A] = 0.5 [Ao]

       substituting the values in (1) , time = 0.5/0.0032 = 156.25 seconds

d) from a , Ao = 0.02136 M , A = Ao/2= 0.01068

                       1/0.01068 M = 1/0.02136 M + 0.0032 M-1s-1 x t

              t = 156.25 sec

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.