Exercise 13.50: Problems by Topic The Integrated Rate Law and Half Life The foll
ID: 487929 • Letter: E
Question
Exercise 13.50: Problems by Topic The Integrated Rate Law and Half Life The following reaction was monitored as a function of time AB- A B A plot of 1/IABl versus time yields a straight line with slope 5.6x10 2 (M s) What is the half-life when the initial concentration is 0.56 M? Express your answer using two significant figures. Submit Mr Answers Glve up Incorrect Try Again, 9 attempts remaining Part D If the initial concentration of AB is 0.260 M, and the reaction mixture initially contains no products, what are the concentrations of A and Bafter 80 s Enter your answers numerically separated by a comma. Express your answers using two significant figures.Explanation / Answer
1.The integrated law of second order reaction is
1/[A]t =kt + 1/[A]0
where, [A]t = Concentration at time t
[A]0 = Initial concentration
k = rate constant
this equation in the form of straight line y = mx + C
where, slope ,m = k =5.6×10^-2M^-1.s^-1
initial concentration is = 0.56M
a) Half life, t1/2 = 1/k[A]0
= 1/5.6×10^-2(M^-1.s^-1)×0.56M
= 32seconds
b) initial concentration [A]0
1/[A]t =5.6 × 10^-2(M^-1s-1) × 80s + 1/0.260M
[A]t = 0.12M
2)Integrated first order equation
ln[A]t = -kt + ln[A]0
where, [A]0 and [A]t are concentration at initial and at time t
k = Rate constant
This equation is in the form of straight line y= mx + C
Therefore , Slope = - k = - 4.4×10^-3/s
k =4.4×10^-3s^-1
a) Half life of first order , t1/2= 0.693/k
t1/2= 0.693/4.4×10^-3s^-1
= 158 seconds
b) Initial concentration [A]0 =0.260M
time = 230seconds
k = 4.4 × 10^-3/s
ln[A]t = - 4.4 × 10^-3s^-1× 230s + ln(0.260M)
2.303log[A]t = -4.4×10^-3s^-1 × 230s + 2.303log(0.260)
[A]t = 0.095M
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