A reaction A ---> B has the following time dependence for the concentration of [

Question

A reaction A ---> B has the following time dependence for the concentration of [A] vs time. For t=(0 s, 5 s, 10 s, 15 s, 25 s) the concentration of [A]=(30.00 M, 23.48 M, 18.38 M, 14.39 M, 8.81 M). The initial concentration of [A] is the value at t=0 s. (A)Calculate the values of the rate constant k assuming that the reaction is first order.

What is the value of k at 5 s?

What is the value of k at 10 s?

What is the value of k at 15 s?

What is the value of k at 25 s?

(B)Calculate the value of k assuming that the reaction is second order.

What is the value of k at 5 s?

What is the value of k at 10 s?

What is the value of k at 15 s?

What is the value of k at 25 s?

(C)Use your results from parts A and B above to decide what the order of the reaction is for your data. (Enter 1 or 2)

(D)What is the concentration of [A] at time t=55 s?

Explanation / Answer

For 1st order reaction,

lnCA= lnCAO-kt

K=( lnCAO- lnCA)/t

Where CA= Concentration of A at any time, CAO= initial concentration

at 5 seconds. K = ( ln30-ln23.48)/5 = 0.049/sec

at 10 seconds =(ln30-ln18.38)/10 = K 0.049/sec

at 15 seconds, K= (ln30-ln14.39)/15=0.049/sec

at 25 seconds, K= (ln30-ln8.81)/25= 0.049/sec

for 2nd order reaction

1/CA= 1/CAO+Kt

1/CA-1/CAO= Kt

K= 1/t*(1/CA-1/CAO)

Where CA= Concentration of A at any time, CAO= initial concentration

At 5 seconds , K= (1/5)*{1/23.48-1/30)=0.001851/M.sec

At 10 seconds, K= (1/10)*(1/18.38-1/30)= 0.002107/M.sec

At 15 seconds, K= (1/15)*(1/14.39-1/30)= 0.002401/M.sec

At 25 seconds, K= (1/25)*(1/8.81-1/30)= 0.003207 /M.sec

Since the rate constant should not vary with time, it is not second order. From the 1st order data, where the rate constnat is same at 0.049 /sec the reaction is 1st order.

For 1st order reaction, lnCA= lnCAO-Kt

At 55 seconds,

lnCA= ln30-0.049*55, CA= 2.03 M

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