Question

The rate constant for this second‑order reaction is 0.670 M − 1 ⋅ s − 1 at 300 ∘ C. A ⟶ products How long, in seconds, would it take for the concentration of A to decrease from 0.690 M to 0.360 M?

Answer 1

To determine the time it takes for the concentration of A to decrease from 0.690 M to 0.360 M in a second-order reaction, we can use the integrated rate law for a second-order reaction. The time would be approximately 3.43 seconds.

Step-by-step explanation:

To determine the time it takes for the concentration of A to decrease from 0.690 M to 0.360 M in a second-order reaction, we can use the integrated rate law for a second-order reaction. The integrated rate law is given by: 1/[A] = kt + 1/[A]₀, where [A] is the concentration at time t, [A]₀ is the initial concentration, k is the rate constant, and t is the time. Rearranging the equation and plugging in the values, we get: t = (1/[A] - 1/[A]₀) / k. Substituting the given values, we have: t = (1/0.360 M - 1/0.690 M) / 0.670 M⁻¹⋅s⁻¹ = 3.43 s. Therefore, it would take approximately 3.43 seconds for the concentration of A to decrease from 0.690 M to 0.360 M.

Answer 2

t = 1.983 s

Step-by-step explanation:

• A → Pdts
• - rA = K (CA)∧α = - δCA/δt

∴ α: order = 2

∴ K: rate constant = 0.670/M.s.....T = 300°C

⇒ - δCA/δt = (0.670/M.s)(CA²)

⇒ - ∫δCA/CA² = (0.670)∫δt

⇒ [ 1/CA - 1/CAo ] = 0.670*t

∴ CA = 0.360 M

∴ CAo = 0.690 M

⇒ [ 1/0.360 - 1/0.690 ] = 0.670*t

⇒ [ 1.3285/M ] = (0.670/M.s)*t

⇒ t = 1.983 s