Aspirin, C9Hg04, slowly decomposes at room temperature by reacting with water in

Question

Aspirin, C9Hg04, slowly decomposes at room temperature by reacting with water in the atmosphere to produce acetic acid, HC2H302, and 2-hydroxybenzoic acid, C7H603 (this is why old bottles of aspirin often smell like vinegar): C9H804+ H2oyT; HC2H302+ C7H603 Concentration and rate data for this reaction are given below. Determine the value of the rate constant (k) for this reaction. 0.0200 0.0100 0.0100 0.0200 2.4 10-13 9.6 10-13 4.8 10-13 0.0800 0.0200 Remember that answers in scientific notation can be written as 3e-4

Explanation / Answer

Consider a rate law for the given reaction

Rate = K[C9H8O4]x[H2O]y ----------(1)

Where K = Rate constant, x is order of reaction with respect to C9H8O4 and y is order of reaction with respect to H2O

From Data, [C9H8O4] = 0.0100 M, [H2O] = 0.0200 M,

Rate = 2.4*10^-13 M/s

From (1) we can write

2.4 * 10^-13 M/s = K(0.0100 M)x(0.0200M)y ------(2)

From data ,

[C9H8O4] = 0.0100 M, [H2O] = 0.0800 M, Rate = 9.6*10^-13

From (1) 9.6*10^-13 M/s = K(0.0100M)x(0.0800M)y -----(3)

Divide (3) by (2)

(9.6*10^-13 M/s)/(2.4*10^-13 M/s) = {K(0.0100 M)x(0.080M)y} /{K(0.010M)x(0.020M)y}

4 = 4^y

y= 1, it means order of reaction with respect to H2O is 1

From data

[C9H8O4] = 0.0200 M , [H2O] = 0.0200 M, Rate = 4.8*10^-13 M/s

From (1)

4.8*10^-13 M/s = K(0.0200M)x(0.0200M)y ------(4)

Divide (4) by (2)

(4.8*10^-13 M/s)/(2.4*10^-13 M/s) = {(0.0200M)x(0.0200M)y} /{(0.010M)x(0.020M)y}

2 = 2^x

x = 1, it means order of reaction with respect to C9H8O4 is 1

Put all the values in (2) to calculate rate constant

2.4*10^-13 M/s = K(0.010M)(0.020M)

2.4*10^-13 M/s = K*0.0002M^2

K = (2.4*10^-13 M/s) / 0.0002 M^2

K = 12 * 10^-10 M^-1s^-1

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