The second-order rate constant for the following gas-phase reaction is . We star

Question

The second-order rate constant for the following gas-phase reaction is . We start with 0.495 mol  in a 3.50-liter container, with no  initially present.

The second-order rate constant for the following gas-phase reaction is 0.0395 M1.s1. We start with 0.495 mol CoF, in a 3.50-iter container, with no C4Fs initially present. a. What will be the concentration of C2F4 after 2.50hours? Concentration What will be the concentration of b. C4Fs after 2.50 hours? Concentration c. What is the half-life of the reaction for the initial C2F4 concentration given in part (a)? Half-life d. How long will it take for half of the C2F4 that remains after 2.50hours to disappear? s or hr

Explanation / Answer

second-order reaction

rate constant k = 0.0395 M-1 s-1

Initial concentration of C2F4 = moles/Volume = 0.495/3.5

[A0] = 0.1414 M

Part a

Time t = 2.50 hr x 3600s/hr = 9000 s

Final concentration [A] =?

1/[A] = 1/[A0] + kt

= (1/0.1414) + 0.0395*9000

1/[A] = 362.57

[A] = 0.00275 M

Final concentration of C2F4 = 2.75*10^-3 M

Part b

Moles of C2F4 after 2.5 hr = 2.75 x 10^-3 M x 3.5 L

= 9.65 x 10^-3

Moles consumed = 0.495 - 9.65 x 10^-3 = 0.4853 mol

Moles of C4F8 produced

= 1 mol C4F8 x 0.4853 mol C2F4 / 2 mol C2F4

= 0.2427 mol

concentration of C4F8 after 2.50 hr = 0.2427 mol/3.5L

= 0.06933 M = 6.93 x 10^-2 M

Part c

Half life for second order reaction

t1/2 = 1/k[A0] = 1/(0.0395*0.1414 )

= 0.0055853 = 5.585 x 10^-3 s

Part d

C2F4 remains after 2.50 hr

[A0] = 2.75*10^-3 M

[A] = 2.75*10^-3/2 = 0.001375 M

1/[A] = 1/[A0] + kt

1/0.001375 = 1/2.75*10^-3 + 0.0395*t

t = 9205.98 s

t = 2.56 hr

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