Question

Q-4 (7.10). The concentration of a reactant in a first-order chemical reaction that proceeds at a rate k can be described as follows: In(C) = ln(Co) - kt, where C is the concentration of the reactant at time t, Co is the initial concentration and t is the elapsed time since the reaction started. Consider an initial concentration of Co = 0.3 mol/L. The experiment was repeated n times to give a geometric mean of the concentration at time t = 450 seconds of 0.22 mol/L. The geometric standard deviation of the concentration at time t = 450 seconds is 1.17. (a) Compute the mean of the rate constant k. (b) Compute the standard deviation of the rate constant k. Hint: Use the fact that k is a linear function of y = In (C).

Answer 1

Answer:

Given that:

The experiment was repeated n times to give a geometric mean of the concentration at time t = 450 seconds of 0.22 mol/L.

a)

here,

InC = inCo - kt

Here

Co = 0.3mol/L

let's say n samples are

C1, C2, C3....cn

let say geometric mean

Cg = (CC ....C)1/n

so, ln  

Cg = InC1 +InC2+ .....InCnln

here for n times geometric mean for t = 450 seconds ; Cg = 0.22 mol/ L

so mean rate constant k =

k = InCo - Incal/t = In0.30 – In0.22/450

k=6.892 * 10-4sec-1

Mean of rate constant

ko = 6.892 * 10-4 sec-1

(b) let say geometric mean standard deviation of concentration is 1.17

Sg = exp[1/(n-1) In(Ci/Cg)?]

so here sg = 1.17

In(C;/C)2 = ln(1.17)(n − 1)

so as k is linear function of ln C .

so the

Sk = 1/(n − 1) (ki – k0)

Sk = koln(1.17)

=

6.892 * 10-sec-1 *In(1.17)

= 1.082 * 10-4 sec-1