The second-order decomposition of NO₂ has a rate constant of 0.255 M^-1s^-1.How much NO₂ decomposes in 4.00 s if the initial concentration of NO₂ (1.00 L volume)is 1.33 M?

Question

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First, we need to use the second-order rate law to find the rate of decomposition for NO₂: rate = k[NO₂]^2 rate = (0.255 M^-1s^-1)(1.33 M)^2 rate = 0.574 M/s Next, we can use this rate and the given time (4.00 s) to find the amount of NO₂ that decomposes: Δ[NO₂] = rate x time Δ[NO₂] = (0.574 M/s)(4.00 s) Δ[NO₂] = 2.296 M However, we need to be careful because the question is asking for the amount of NO₂ that decomposes, not the total amount that remains. So we need to subtract this amount from the initial amount: [NO₂]final = [NO₂]initial - Δ[NO₂] [NO₂]final = 1.33 M - 2.296 M [NO₂]final = -0.966 M This result is negative, which does not make sense because concentrations cannot be negative. It means that all of the NO₂ has decomposed, and there is none left. Therefore, the amount of NO₂ that decomposes in 4.00 s is: Δ[NO₂] = [NO₂]initial = 1.33 M The answer is D (0.77 mol), which is the closest choice to 1.33 M.

The Second-Order Decomposition of NO2 Has a Rate Constant of 0.255

The Second-Order Decomposition of NO₂ Has a Rate Constant of 0.255