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Aspirin, C9H8O4, Slowly Decomposes at Room Temperature by Reacting with Water

Question 79

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Aspirin, C9H8O4, slowly decomposes at room temperature by reacting with water in the atmosphere to produce acetic acid, HC2H3O2, and 2-hydroxybenzoic acid, C7H6O3 (this is why old bottles of aspirin often smell like vinegar) : C9H8O4 + H2O \rarr HC2H3O2 + C7H6O3
Concentration and rate data for this reaction are given below. [C9H5O4](M) [H2O](M) Rate(M/s) 0.01000.02002.4×10130.01000.08009.6×10130.02000.02004.8×1013\begin{array}{ccc}\mathbf{\left[\mathrm{C}_{9} \mathrm{H}_{5} \mathrm{O}_{4}\right](\mathrm{M}) } & \mathbf{\left[\mathrm{H}_{2} \mathrm{O}\right] (M) } & \mathbf{Rate (M/s) } \\\hline 0.0100 & 0.0200 & 2.4 \times 10^{-13} \\0.0100 & 0.0800 & 9.6 \times 10^{-13} \\0.0200 & 0.0200 & 4.8 \times 10^{-13} \\\end{array} Write the rate law for this reaction and calculate k (be sure to include the correct units) .


A) rate = k[C9H8O4][H2O]2, and the rate constant is 1.2 *10-9 M-2s-1 or 1.2 * 10-9 1/(M2 s)
B) rate = k[C9H8O4], and the rate constant is 1.2 * 10-9 -1 or 1.2 * 10-9 1/ s
C) rate = k[C9H8O4]2[H2O], and the rate constant is 1.2 * 10-9 M-2s-1 or 1.2 * 10-9 1/(M2 s)
D) rate = k[C9H8O4][H2O], and the rate constant is 1.2 *10-9 M-1s-1 or 1.2 * 10-9 1/(M s)
E) None of the above

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