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View History Bookmarks Window Help 88% . Sat 10:04 PM a chegg.com CH 09 HW-1 ± Half-life for First and Second Order Reactions 3 of 15> A certain first-order reaction (A-+products) has a rate constant of 6.60x103s-1 at 45 C. How many minutes does it take for the concentration of the reactant, IA), to drop to 625% of the original The half-ife of a reaction, t1/2, is the time it takes for the reactant concentration [A to decrease by half. For example, after one haltf-life the concentration falls from the initial concentration Express your answer with the appropriate units View Available Hint(s) Alo to A)o/2, after a second half-life to |Al /4 afer a third haif-ife to [Alo/8, and so on. on TAi alue Units Submit Previous Answers Request Answer Incorrect; Try Again; 4 attempts remaining: no points deducted The compound unit you entered is not recognized PartB Acertain second-order reaction (B+products) has a rate constant of 1.60 10-3M at 27 C and an initial half-ife of 216s.What is the concentration of the reactant B after one haif-life? Express your answer with the appropriate units View Available Hint(s)

Explanation / Answer

Part A

Answer

7 minute

Explanation

Integrated first order equation is

ln[A]t= - kt + ln[A]0

where,

  [A]t = Concentration at time t, 6.25%

[A]0 = Initial concentration,100%

k = rate constant, 6.60×10-3s-1

rearranging the equation

2.303log([A]0/[A]t) = kt

substitute the value

2.303log(100/6.25) = 6.60×10-3s-1 × t

t = 420.16s=7minutes

Part B

Answer

1.447M

Explanation

For second order reaction

half life, t1/2 = 1/k[A]0

[A]0 = 1/(t1/2 × k) = 1/(216s × 1.60×10-3 M-1s-1) = 2.894M

So, the initial concentration is = 2.894M

Concentration after one half life = 2.894M/2 = 1.447M

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