Show that if a first order reaction runs for 10 half-lives, only 0.098% of the o
ID: 1004306 • Letter: S
Question
Show that if a first order reaction runs for 10 half-lives, only 0.098% of the original starting material remains. What general type of reaction is the hydrolysis of t-butylchloride? Derive the expression that relates rate constant to the half-life.. Substitute this information into the equation for a first-order reaction. In A_0/A = kt Do the units you use to measure the volume of base effect your calculated value of k? Why or why not? Referring to the equations for the treatment of your data, if your were to plot ln(V_infinity/V_infinity - V_t) vs. t, what would the slope equal?Explanation / Answer
1) let the half life = t1/2 minutes
Ten half lives means = 10t1/2 minutes = t minutes
now we know that K = 0.693 / t1/2
lnA0/ A = Kt
lnA0/A = 0.693 X t / t1/2
t = 10 X t1/2
So lnA0/A = 0.693 X 10 t1/2 / t1/2
ln A0/ A = 6.93
Taking antilog
A0/ A = 1022.49
A = A0 / 1022.49 = 9.78 X 10^-4
A = 9.78X 10^-4 A0
A = 0.0978 % = 0.098%
2) the general type of reaction of t-butyl chloride is SN1 reaction
3) We know that
Ln (A0 / A) = kt
Let us put A = A0/ 2
ln (2A0 / A0) = kt
2.303 X log (2) = kt
k = 0.693 / t
t = half life
3) no it will not affect the calculate value of k as the unit of K is time^-1
It is independant of volume unit (the unit gets cacelled in the rate expression)
4) if we plot a graph then the slope will be equal to
Slope = 2.303 / K