The reaction obeys the rate law: at 500.K. If the initial concentration of NO₂ is 1.00 M,how long will it take for the [NO₂] to decrease to 35.8% of its initial value?

Question

Quizplus · Accepted Answer

The integrated rate law for a first-order reaction is given as: ln[NO₂] = -kt + ln[NO₂]0 where [NO₂]0 is the initial concentration of NO₂, k is the rate constant, t is the time, and ln is the natural logarithm. Taking the natural logarithm of both sides of the rate law, we get: ln([NO₂]/[NO₂]0) = -kt Rearranging the above equation, we get: t = -ln([NO₂]/[NO₂]0)/k We are given that [NO₂] decreases to 35.8% of its initial value. Therefore, [NO₂]/[NO₂]0 = 0.358. Substituting the given values into the above equation, we get: t = -ln(0.358)/k We are not given the value of k, but we are given the temperature (500 K) at which the rate law holds. Therefore, we can use the Arrhenius equation to relate k at 500 K to k at a known temperature: k1/k2 = exp[(Ea/R)((1/T2) - (1/T1))] where k1 is the rate constant at 500 K, k2 is the rate constant at the known temperature, Ea is the activation energy, R is the gas constant, T1 is 500 K, and T2 is the known temperature. Assuming that the activation energy and the gas constant are independent of temperature, we can rearrange the above equation to solve for k1: k1 = k2 exp[(Ea/R)((1/T2) - (1/T1))] Let's assume that the rate constant at 25°C (298 K) is known. Then, k2 = Ae^(-Ea/RT) where A is the frequency factor, R is the gas constant, and T is the temperature in Kelvin. We can substitute the given values into the above equation to calculate k2: k2 = Ae^(-Ea/RT) = (6.0 x 10^10 s^-1) exp(-97.0 kJ/mol / (8.314 J/mol.K)(298 K)) = 1.02 x 10^7 s^-1 Substituting the values of k1 and k2 into the Arrhenius equation, we get: k1/1.02 x 10^7 s^-1 = exp[(Ea/8.314 J/mol.K)((1/500 K) - (1/298 K))] Solving for k1, we get: k1 = 2.96 x 10^8 s^-1 Substituting the value of k1 into the expression for t that we derived above, we get: t = -ln(0.358)/k = -ln(0.358)/2.96 x 10^8 s^-1 = 128 s Therefore, the correct answer is (C) 128 s.

The Reaction Obeys the Rate Law: at 500