Question

In an experiment studying the decomposition of acetaldehyde, a half-life of 328 seconds is obtained when the starting concentration is 9.72 x 10-3 M. For each of the assumptions below, calculate the reaction’s rate constant and the concentration of acetaldehyde after the reaction has been going for 10 minutes, 15 minutes and 20 minutes.

a. Assuming the reaction is 0th order

b. Assuming the reaction is 1st order

c. Assuming the reaction is 2nd order

Answer 1

the relation between rate (-dA/dt), where A= concentration of acetaldehyde is = K[A]ⁿ, where K= rate constant and n is order.

whe n=0, the reaction is zero order, -dA/dt= K, when integrated at t= 0, [A]= [A]0, at t=t, [A]= [A]

[A]= [A]o-Kt (1), given [A]o = 9.72*10^-3 M

given half life is the time required for the concentration to drop to 50% of the initial value,

that is [A]= [Ao]/2, at t= 328 seconds

K*328= 9.72*10^-3*(1-0.5)

K= 1.48*10^-5/M.sec

when t= 10min, t= 10*60 sec= 600sec,   from Eq.1, [A] = 9.72*10^-3- 1.48*10^-5*600 =0.000829 M

when t= 15 min, t= 15*60 =900 sec, from Eq.1, [A] = 9.72*10^-3- 1.48*10^-5*900=-ve , the reaction is complete since [A] cannot be negative, hence at 20 minutes also the reaction is complete.

2. for 1st order reaction, n=1 and the rate becomes -dA/dt= K[A], when integrated, lnA= ln[A]0-Kt (2)

when [A]= [A]o/2, the half life is t= 328 sec

K= 0.693/328=0.00211 /sec

at t= 10 min= 10*60 sec= 600 sec, Eq.2 becomes ln [A]= ln(9.72*10^-3)-0.00211*600 , [A] =0.002741M

at t=15min= 15*60sec= 900sec, Eq.2 becomes ln[A] =ln(9.72*10^-3)-0.00211*900, [A] =0.001455

at t= 20min =20*60 sec= 1200sec , Eq.2 becomes ln[A]= ln(9.72*10-3)- 0.00211*1200, [A]= 0.000773

3. for n=2, the rate expression becomes -dA/dt= K[A]2, when integrated, 1/[A] = 1/[AO]+ Kt (3)

when [A]= [A]0/2 , Eq.3 becomes 1/[Ao]2= 1/[A]O+K*328

1/[A]o= K*328, K= 1/(9.72*10^-3*328)= 0.313/M.sec

when t= 10 min, t=10*60=600sec, Eq. 3 becomes 1/A= 1/(9.72*10^-3)+ 0.313*600 , [A] = 0.00344M

at t= 15 min, t= 15*60= 900sec, Eq.3 becomes 1/A = 1/(9.72*10^-3)+0.313*900 , [A]=0.0026M

at t=20 min, t= 20*60sec = 1200 sec, Eq.3 becomes, 1/A = 1/(9.72*10^-3)+ 0.313*2000, [A]=0.001372