The first-order rate constant for the decomposition of $\mathrm{N}_{2} \mathrm{O}_{5}, 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} .(\mathbf{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

(a) Moles of N2O5 remaining after 5 minutes

We are given the initial concentration and the rate constant for the decomposition of N2O5. We will use the first-order rate law equation to calculate the remaining amount of N2O5 after 5 minutes: \[a_t = a_0e^{-kt}\] where: - \(a_t\) is the amount of N2O5 at time t - \(a_0\) is the initial amount of N2O5 - k is the first-order rate constant (6.82 x 10^-3 s^-1) - t is the time (5 minutes = 300 seconds) Plugging the values into the equation, we have: \[a_t = (0.0250) e^{-(6.82 \times 10^{-3})(300)}\] Calculating the result, we get: \[a_t = 0.0190 \,\text{mol}\] Hence, there will be 0.0190 moles of N2O5 remaining after 5 minutes.

02

(b) Time to drop the quantity of N2O5 to 0.010 mol

We will use the same first-order rate law equation to find the time it takes for the amount of N2O5 to reach 0.010 mol. Rearranging the equation to solve for time: \[t = -\frac{\ln(\frac{a_t}{a_0})}{k}\] Plugging the values into the equation, we have: \[t = -\frac{\ln(\frac{0.010}{0.0250})}{6.82 \times 10^{-3}}\] Calculating the result, we get: \[t = 445 \,\text{seconds}\] Converting the time to minutes, we have: \[\text{Time} = \frac{445}{60} \, \text{min}\] Hence, it will take approximately 7.42 minutes for the quantity of N2O5 to drop to 0.010 mol.

03

(c) Half-life of N2O5 at 70°C

The half-life of a first-order reaction can be determined using the following equation: \[t_{1/2} = \frac{0.693}{k}\] We already have the rate constant, so plugging the value into the equation, we get: \[t_{1/2} = \frac{0.693}{6.82 \times 10^{-3}}\] Calculating the result, we get: \[t_{1/2} = 102 \, \text{seconds}\] Converting this to minutes, we have: \[\text{Half-Life} = \frac{102}{60} \, \text{min}\] Hence, the half-life of N2O5 at 70°C is approximately 1.7 minutes.

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