The decomposition of \(\mathrm{A}_{2} \mathrm{~B}_{2}\) to \(\mathrm{A}_{2}\) and \(\mathrm{B}_{2}\) at \(38^{\circ} \mathrm{C}\) was monitored as a function of time. A plot of \(1 /\left[\mathrm{A}_{2} \mathrm{~B}_{2}\right]\) vs. time is linear, with slope \(0.137 / M \cdot \mathrm{min}\) (a) Write the rate expression for the reaction. (b) What is the rate constant for the decomposition at \(38^{\circ} \mathrm{C} ?\) (c) What is the half-life of the decomposition when \(\left[\mathrm{A}_{2} \mathrm{~B}_{2}\right]\) is \(0.631 \mathrm{M} ?\) (d) What is the rate of the decomposition when \(\left[\mathrm{A}_{2} \mathrm{~B}_{2}\right]\) is \(0.219 \mathrm{M}\) ? (e) If the initial concentration of \(\mathrm{A}_{2} \mathrm{~B}_{2}\) is \(0.822 \mathrm{M}\) with no products present, then what is the concentration of \(\mathrm{A}_{2}\) after \(8.6\) minutes?

Step by step solution

01

(a) Determine the Rate Expression

As a plot of \(1/\left[\mathrm{A}_{2}\mathrm{~B}_{2}\right]\) vs. time is linear, it indicates that the reaction is first order with respect to A₂B₂. Let's write the rate of the reaction: Rate = k[A2B2]

02

(b) Find the Rate Constant

To find the rate constant, we'll use the slope given in the problem. Since the plot of \(1/\left[\mathrm{A}_{2}\mathrm{~B}_{2}\right]\) vs. time is linear, the slope of this plot (\(m\)) is equivalent to the rate constant for the decomposition (k): k = 0.137 M⁻¹·min⁻¹

03

(c) Calculate the Half-life

For a first-order reaction, the half-life (t₁/₂) is given by the formula: \(t_{1/2} = \frac{0.693}{k}\) We can plug in the rate constant found earlier: \(t_{1/2} = \frac{0.693}{0.137\thinspace M^{-1} \cdot min^{-1}} \approx 5.06\thinspace min\)

04

(d) Find the Rate of Decomposition

To find the rate of decomposition when [A₂B₂] is 0.219 M, we can use the rate expression determined earlier: Rate = k[A₂B₂] = (0.137 M⁻¹·min⁻¹)(0.219 M) ≈ 0.030\thinspace M·min⁻¹

05

(e) Calculate the Concentration of A₂

To find the concentration of A₂ after 8.6 minutes, we'll use the integrated rate law for a first-order reaction: \(\left[\mathrm{A}_{2}\right] = \left[\mathrm{A}_{2}\mathrm{~B}_{2}\right]_{0} \cdot \left(1 - e^{-kt}\right)\) Let's plug in the values for the initial concentration of A₂B₂ (0.822 M), the rate constant (0.137 M⁻¹·min⁻¹), and time (8.6 min): \(\left[\mathrm{A}_{2}\right] = 0.822\mathrm{M} \cdot \left(1 - e^{-0.137\cdot8.6\mathrm{M^{-1}\cdot min^{-1}\cdot min}}\right) \approx 0.402\thinspace M\) Therefore, the concentration of A₂ after 8.6 minutes is approximately 0.402 M.

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