(a) The reaction $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+\( \)\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)$ is first order with in \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)\) and zero-order in \(\mathrm{H}_{2} \mathrm{O}\). At \(300 \mathrm{~K}\) the rate constant equals \(3.30 \times 10^{-2} \mathrm{~min}^{-1} .\) Calculate the half- life at this temperature. \((\mathbf{b})\) If the activation energy for this reaction is \(80.0 \mathrm{~kJ} / \mathrm{mol}\), at what temperature would the reaction rate be doubled?

In a first-order reaction, the integrated rate equation is given by the formula: \[ ln\frac{[A]_{0}}{[A]_{t}} = kt. \] To find the half-life (t₁/₂), we set \([A]_{t}\) as \(\frac{[A]_{0}}{2}\), since half the reactant remains at the half-life time. Plugging this into the formula, we get: \[ ln\frac{[A]_{0}}{\frac{[A]_{0}}{2}} = kt_{1/2}. \] Now, we are left with: \[ ln2 = kt_{1/2}. \] Solving for t₁/₂, we have: \[ t_{1/2} = \frac{ln2}{k}. \] The rate constant at 300 K is 3.30 x 10⁻² min⁻¹. Therefore, we can calculate the half-life as: \[ t_{1/2} = \frac{ln2}{3.30 × 10^{-2} \,\text{min}^{-1}}. \] Calculating, we get the half-life: \[ t_{1/2} \approx 21.0\, \text{min}. \]

To find the temperature at which the reaction rate doubles, we will use the Arrhenius equation, which is given by: \[ k = Ae^{-\frac{E_{a}}{RT}}, \] where k is the rate constant, A is the pre-exponential factor, Eₐ is the activation energy, R is the gas constant, and T is the temperature in Kelvin. Since the rate doubles, we can write another equation for the new rate constant, k': \[ k' = Ae^{-\frac{E_{a}}{R(T')}}, \] where k' is twice the original rate constant and T' is the new temperature. Now, we will divide the second equation by the first one: \[ \frac{k'}{k} = \frac{Ae^{-\frac{E_{a}}{R(T')}}}{Ae^{-\frac{E_{a}}{RT}}} = \frac{e^{-\frac{E_{a}}{R(T')}}}{e^{-\frac{E_{a}}{RT}}}. \] Since k' = 2k, we can rewrite the equation as: \[ 2 = \frac{e^{-\frac{E_{a}}{R(T')}}}{e^{-\frac{E_{a}}{RT}}}. \] Applying ln on both sides, we get: \[ ln2 = \frac{E_{a}}{RT} - \frac{E_{a}}{R(T')}. \] We are given Eₐ as 80.0 kJ/mol = 80000 J/mol. Also, R = 8.314 J/mol·K. We can now solve for T': \[ T' = \frac{E_{a}RT}{R(E_{a} - ln2RT)} = \frac{80000 × 8.314 × 300}{8.314(80000 - ln2 × 8.314 × 300)}. \] Calculating T', we find the new temperature: \[ T' \approx 307.2\, K. \] In conclusion, the half-life of the reaction at 300 K is approximately 21.0 minutes, and the reaction rate would double at a temperature of about 307.2 K.