Algebra Question

Question

Reacting with water in an acidic solution at a particular​ temperature, compound A decomposes into compounds B and C according to the law of uninhibited decay. An initial amount of .40 M of compound A decomposes to .37 M in 30 minutes. How much of compound A will remain after 2 ​hours? How long will it take until 0.10 M of compound A​ remains?

Answer 1

Uninhibited decay generally implies first order decay which follows A_t = A_oe^-kt where A_t is amount remaining at time t, A_o is the initial amount, k is the rate constant and t is the time. Another way to view this for first order reactions is that F = 0.5ⁿ where F is the fraction remaining and n is the number of half lives that have elapsed.

rate = 0.40 M - 0.37 M/30 min = 0.001 M/min. Since it is first order, one can write the rate equation as rate = k[A]

0.001 M/min = k[0.4 M]

k = 0.0025 min^-1

half life for first orders reaction = 0.693/k = 277.2 min

2 hours = 120 min and thus 277.2 min/120 min = 2.31 half lives have elapsed

F = 0.5^2.31 = 0.2017 = fraction remaining

[A] remaining after 2 hours = 0.40 M x 0.2017 = 0.081 M

How long will it take until 0.1M remains? 0.1M/0.4M = 0.25 fraction remaining

0.25 = 0.5ⁿ and solving for n we obtain 2 half lives and 2 half lives = 2 x 277.2 min = 554.4 min = 9.24 hours

You can also solve this using the formula above of At = A_oe^-kt after solving for k as above