Question

For a first-order reaction, the half-life is constant. It depends only on the rate constant k and not on the reactant concentration. It is expressed as

t1/2=0.693k

For a second-order reaction, the half-life depends on the rate constant and the concentration of the reactant and so is expressed as

t1/2=1k[A]0

Part A

A certain first-order reaction (A→products ) has a rate constant of 5.10×10⁻³ s−1 at 45 ∘C . How many minutes does it take for the concentration of the reactant, [A] , to drop to 6.25% of the original concentration?

Express your answer with the appropriate units.

Part B

A certain second-order reaction (B→products ) has a rate constant of 1.05×10⁻³M−1⋅s−1 at 27 ∘C and an initial half-life of 224 s . What is the concentration of the reactant B after one half-life?

Express your answer with the appropriate units.

Answer 1

Part A:

Conventration after every consecutive half life decreases by a factor of 2. That is if initial concentration is C then after 1 half life Concentration becomes C/2, after 2 half lives it further decreases by a factor of 2 and becomes C/2^{2 _,} after three half lives it becomes C/2³ and so on.

Now for concentration to become C×6.25/100, we need to solve this equation for n : C×6.25/100 = C/2ⁿ which gives

2^{​​​​​​n} =100/6.25 = 16 = 2⁴, so n=4. It takes 4 half lives to drop concentration to 6.25% of initial. That means total time taken

= 4×t_1/2 = 4×(ln2)/k = 4×(ln2)/(5.10×10^-3 s^-1) = 543.6 s

= 9.06 minutes

Part B:

t_1/2 = 1/(k[A]_o) using it we have 224 s= 1/(1.05×10^-3M^-1s^-1×[A]_o), solving it we get [A]_o = 4.25 M. After one half life concentration becomes half of initial. So concentration would be (4.25/2) M = 2.125 M = 2.12 M.

Comment in case of any doubt.