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A gas-phase reaction A ⇌ B is known to have a rate constant of 0.05 L/mol/s at 300 K. If the initial concentration of A is 0.1 M and the system reaches equilibrium after 2 minutes, what is the concentration of B at equilibrium? (Assume constant pressure and ideal gas behavior)

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To solve this problem, we need to use the integrated rate law for a first-order reaction and the equilibrium constant expression.First, let's find the integrated rate law for a first-order reaction:[A]t = [A]0 * e^-kt where [A]t is the concentration of A at time t, [A]0 is the initial concentration of A, k is the rate constant, and t is the time.Since the system reaches equilibrium after 2 minutes, we can plug in the values:[A]t = 0.1 * e^-0.05 * 2 * 60 [A]t = 0.1 * e^-6 [A]t  0.1 * 0.002478752 = 0.0002478752 MNow, let's find the concentration of B at equilibrium. Since the reaction is A  B, the change in the concentration of A is equal to the change in the concentration of B. Therefore, the concentration of B at equilibrium can be found by subtracting the equilibrium concentration of A from the initial concentration of A:[B]eq = [A]0 - [A]t[B]eq = 0.1 - 0.0002478752[B]eq  0.0997521248 MSo, the concentration of B at equilibrium is approximately 0.0998 M.

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