The reaction AB(aq) → A(g) + B(g) is second order in AB and has a rate constant of 0.0118 M^-1 s^-1 at 25.0 °C. A reaction vessel initially contains 250.0 mL of 0.100 M AB that is allowed to react to form the gaseous product. The product is collected over water at 25.0 °C. How much time is required to produce 200.0 mL of the products at a barometric pressure of 755.1 mmHg? (The vapor pressure of water at this temperature is 23.8 mmHg.)

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1. Identify the initial concentration of AB, which is 0.100 M, and the initial volume of the solution, which is 250.0 mL.
2. Use the ideal gas law to determine the moles of gas produced. First, calculate the pressure of the gas by subtracting the vapor pressure of water from the barometric pressure: \( P_{gas} = 755.1 \text{ mmHg} - 23.8 \text{ mmHg} \).
3. Convert the volume of the gas produced (200.0 mL) to liters and use the ideal gas law \( PV = nRT \) to find the moles of gas, where \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \) and \( T = 298 \text{ K} \).
4. Determine the change in concentration of AB using the stoichiometry of the reaction and the moles of gas produced. Since the reaction is second order, use the integrated rate law for a second-order reaction: \( \frac{1}{[AB]_t} = \frac{1}{[AB]_0} + kt \).
5. Solve for the time \( t \) using the integrated rate law, where \( k = 0.0118 \text{ M}^{-1} \text{ s}^{-1} \), \([AB]_0 = 0.100 \text{ M}\), and \([AB]_t\) is the concentration of AB after the reaction.