A substance (A) reacts with sodium hydroxide according to the following reaction

Question

A substance (A) reacts with sodium hydroxide according to the following reaction. A + OH^- rightarrow AOH^- r = k[A]^r[OH^-]^y The substance A is blue, the product is colorless, therefore the reaction can be followed by spectrophotometry. Two runs were performed with the initial hydroxide concentrations [OH^-] = 2.00M and [OH^-]_2.0 = 1.00M. For both runs the concentration of A was [A]_a = 1.47 times 10^-2M. The concentration of A, time data for both runs are given below (copy and paste it into Excel for processing). Determine the order of Ain the reaction (x). X =____________Determine the order of [OH] in the reaction (y). y =____________Determine the rate constant, k, of the reaction. k =_________Interpret the effect of [OH]>>[A] in terms of the mechanism of the reaction.

Explanation / Answer

The concentrtion of OH in both trails is significantly higher than the concentration of A. So the reaction is pseudo order.

Hence rate= K[ OH-]m [A]n = K' [A]n

the order is assumed and the firsr orer expression is ln C= lnC0-Kt . When the data of lnC vs t is plotted it is not giving straight line. Hence first order is rejected.

for second order, the expression KAot= XA/(1-XA) when data of XA/(1-XA) vs time is plotted, it is giving a straight line suggesting it to be 2nd order.

The rate constants

Slope is K'CAO= 0.015 , trial-2 = 0.007

for trial-1 CAO= 1.47*10-3, for trial-2, CAO= 1.46*10-3

K' = 0.015/(1.47*10-3)=10.20 for trial-2 K' = 0.007/(1.46*10-3) =4.79

but K= K' [OH-]m

lnK= lnK' +m ln [OH-]

[OH-] = 2M

lnK= ln(10.20) + m ln2 ( trial-1) (1)

lnK= ln (4.79)+ mln1 (Trial-2) (2)

lnK= ln (4.79)

K=4.79

from trial-1, ln (4.79)= ln(10.20)+ m ln2

m=-1

K=4.79

when [OH]> A the reaction becomes pseudo order.

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