The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, \quad 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g), \quad\) at \(\quad 70^{\circ} \mathrm{C}\) is \(6.82 \times 10^{-3} \mathrm{~s}^{-1}\). Suppose we start with \(0.0250 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) in a volume of \(2.0 \mathrm{~L}\). (a) How many moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will remain after \(5.0 \mathrm{~min} ?\) (b) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(0.010 \mathrm{~mol}\) ? (c) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70{ }^{\circ} \mathrm{C} ?\)

Step by step solution

01

Part A: Calculate moles of N2O5 remaining after 5.0 min

Given: Initial moles (A0) = 0.0250 mol Time (t) = 5.0 min = 300 s (converted to seconds) Rate constant (k) = 6.82 × 10^{-3} s^{-1} Use the formula: \[ ln(\frac{[A]_0}{[A]}) = kt \] Rearrange the formula to find the remaining moles (A): \[ [A] = \frac{[A]_0}{e^{kt}} \] Substitute the values: \[ [A] = \frac{0.0250}{e^{(6.82 × 10^{-3})(300)}} \] Calculate the result: \[ [A] = 0.0168 \; \text{mol} \] After 5.0 min, there will be approximately 0.0168 mol of N2O5 remaining.

02

Part B: Calculate time for the quantity of N2O5 to drop to 0.010 mol

Given: Initial moles (A0) = 0.0250 mol Final moles (A) = 0.010 mol Use the formula: \[ ln(\frac{[A]_0}{[A]}) = kt \] Rearrange the formula to find the time (t): \[ t = \frac{ln(\frac{[A]_0}{[A]})}{k} \] Substitute the values: \[ t = \frac{ln(\frac{0.0250}{0.010})}{6.82 × 10^{-3}} \] Calculate the result: \[ t ≈ 348.88 \; \text{s} \] Convert to minutes: \[ time \approx 5.81 \; \text{min} \] It will take approximately 5.81 minutes for the quantity of N2O5 to drop to 0.010 mol.

03

Part C: Calculate the half-life of N2O5 at 70°C

To find the half-life (t1/2) of a first-order reaction, we use the formula: \[ t_{1/2} = \frac{ln(2)}{k} \] Given: Rate constant (k) = 6.82 × 10^{-3} s^{-1} Substitute the value: \[ t_{1/2} = \frac{ln(2)}{6.82 × 10^{-3}} \] Calculate the result: \[ t_{1/2} ≈ 101.73 \; \text{s} \] Convert to minutes: \[ half-life ≈ 1.70 \; \text{min} \] The half-life of N2O5 at 70°C is approximately 1.70 minutes.

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