Verified answer

Part A

Use integrated first order rate law

ln[A] = ln[A]₀ - k∙t

([A] concentration of reactant A at time t, [A]₀ initial reactant concentration, k rate constant)

So the time elapsed unless concentration has dropped from initial [A]₀ to [A] is:

t = (ln[A]₀ - ln[A] ) / k = ln( [A]₀/[A] ) / k

The time takes for the concentration to drop to 6.25% of its initial value, i.e [A] = 0.0625[A]₀; is:

t = ln( 1/0.0625 ) / 5.40×10⁻³ s⁻¹

= 513.4 s

= 8.56 min

Part B

You need to find the initial concentration of the reactant. You can find it from the relation of rate constant and initial concentration to half-life.

Half-life for a second order reaction is given by:

t½ = 1/( k∙[B]₀ )

Hence,

[B]₀ = 1/( k∙t½ ) = 1/( 1.30×10⁻³ M⁻¹∙s⁻¹ ∙ 278s ) = 2.767 M

So after one half-life the concentration is:

[B] = (1/2) ∙ 2.767 M = 1.3835 M

For a first order reaction you can use the equation. At=Ai*e^(-kt) where, At=amount of substance remaining at time t Ai=initial amount of substance k=rate constant t=time Now lets say we started with a 100 g of reactant A. After 68 minutes we only have 54 g left because the reaction is 46% complete. From this we get values to substitute into the equation. At=54 g Ai=100 g k=? t=68 min 54 g=100 g*e^(-k*68 min) 54 g/100 g=e^(-k*68 min) 0.54=e^(-k*68 min) change of base formula ln0.54=-k*68 min k=ln0.54/-68 min=0.0091/min 0.0091=9.1x10^-3 There's your answer