The reaction 2 NO2(g) → 2 NO(g) + O2(g) is a second-order reaction with the following rate law at 298 K:

Rate = (0.543 M–1 s–1)[NO2]2

If the initial concentration of NO2(g) in a closed vessel is 1.50 M at 298 K, what is its concentration (in M) after 42.0 seconds?

Round your answer to FOUR decimal places (i.e. 7.3128) and do NOT include any units.
If your last digit is a zero after a decimal (i.e. 6.40), Carmen may drop the zero. That is OK!

16D

Question

The reaction 2 NO2(g) → 2 NO(g) + O2(g) is a second-order reaction with the following rate law at 298 K:

Rate = (0.543 M–1 s–1)[NO2]2

If the initial concentration of NO2(g) in a closed vessel is 1.50 M at 298 K, what is its concentration (in M) after 42.0 seconds?

Round your answer to FOUR decimal places (i.e. 7.3128) and do NOT include any units.
If your last digit is a zero after a decimal (i.e. 6.40), Carmen may drop the zero. That is OK!

16D

BlackTom AI · Accepted Answer

The problem gives a second-order rate law for the reaction 2 NO2 → 2 NO + O2, with Rate = (0.543 M⁻¹ s⁻¹)[NO2]² and asks for [NO2] after 42.0 s starting from [NO2]₀ = 1.50 M at 298 K.

Since the rate law is second order in NO2, we can use the integrated form for a reaction with rate = k [A]^2: d[A]/dt = -k [A]^2, which integrates to 1/[A] = kt + C, where C is a constant determined by the initial condition.

Apply the initial condition: at t = 0, [NO2] = 1.50 M, so 1/[NO2] = 1/1.50 = 0.6666667 M⁻¹. Therefore C = 0.6666667.

Thus, the concentration at time t is given by 1/[NO2](t) = k t + 0.6666667, with k = 0.543 s⁻¹ M⁻¹.

Calculate kt: k t = 0.543 × 42.0 s = 22.806.
Add the initial constant: 1/[NO2](42.0) = 22.806 + 0.6666667 = 23.4726667 M⁻¹.

Invert to obtain [NO2](42.0): [NO2] = 1 / 23.4726667 ≈ 0.042611 M.

Round to four decimal places as requested: 0.0426. Note that the instruction allows dropping a trailing zero if present, but in this case the value ends with a nonzero digit after rounding, so 0.0426 is the four-decimal result.

Key conceptual check: for a second-order decay with respect to a single reactant, the integrated form 1/[A] = kt + 1/[A]₀ correctly captures the time dependence and yields a decreasing concentration over time, which aligns with the physical expectation for NO2 consumption in this reaction.

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