Question

Calculate the instantaneous rate of decomposition of acetaldehyde when the concentration is 5.55 × 10^{-4} M.

a) 2.63×10−12mol L−1s−1
b) 2.63×10−9mol L−1s−1
c) 2.63×10−5mol L−1s−1
d) 2.63×10−8mol L−1s−1

Answer 1

The calculated instantaneous rate of acetaldehyde decomposition at a concentration of 5.55 × 10-4 M with the given rate constant is 1.45 × 10-14 mol L-1 s-1, which does not match any of the provided options.

Step-by-step explanation:

The decomposition of acetaldehyde (CH3CHO) is a second-order reaction, which means the rate of the reaction depends on the concentration of acetaldehyde squared. The rate law for a second-order reaction is given by the formula rate = k [A]2, where k is the rate constant and [A] is the concentration of the reactant (acetaldehyde in this case).

Given that the rate constant (k) is 4.71 × 10-8 L mol-1 s-1 and the concentration of acetaldehyde ([A]) is 5.55 × 10-4 M, we can calculate the instantaneous rate of decomposition:

rate = (4.71 × 10-8 L mol-1 s-1) × (5.55 × 10-4 M)2
rate = 4.71 × 10-8 × 3.08025 × 10-7 mol2 L-2 s-1
rate = 1.45 × 10-14 mol L-1 s-1

The correct answer is not provided in the options given, indicating a possible mistake in the question or the provided answers.