The concentration of a reactant in a first-order chemical reaction that proceeds

Question

The concentration of a reactant in a first-order chemical reaction that proceeds at a rate k can be described as follows: In C = In Co-kt, where C is the concentration of the reactant at time t, Co is the tial 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 timet 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 = Inc

Explanation / Answer

here,

ln C = ln C0 -kt

Here C0 = 0.3 mol/L

let's say n samples are C1 , C2 ,C 3 ....Cn

let say geometric mean Cg= (C1C2....Cn)1/n

so, ln Cg = [ln C1 + ln C2 + .....ln Cn]/ n

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

so mean rate constant k =

k = [ln C0 - ln Cg] /t = [ln 0.30 - ln 0.22] / 450 = 6.892 x 10-4 sec-1

Mean of rate constant k0 = 6.892 x 10-4 sec-1

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

sg = exp [1/(n-1) ln(Ci /Cg)2 ]

so here sg = 1.17

ln(Ci /Cg)2 = ln (1.17) (n-1)

so as k is linear function of ln C .

so the sk = 1/(n-1) (ki - k0)2

sk = k0 ln (1.17) = 6.892 x 10-4 sec-1 * ln (1.17) = 1.082 x 10-4  sec-1

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