Question

Question 8 (a) (5 points) The reaction A (aq) B (aq) is a first order reaction with respect to A (aq). The half-life of the reaction at 25 C is 377 s. The half-life of the reaction at 35 C is 333 sec. What time will be needed so that the concentration of A (aq) falls from 0.777 M to 0.222 M at 50 C? (b) (3 points) N20 (g) decomposes to give N2 (g) and O2 (g). The reaction is of order one compared to N20 (g). If we start with just N20 (g) in a volume container and fixed temperature and its pressure is 10.00 atm, what will be the total pressure after a half-life? Answers Question 8 a) 507 s b) 12.50 ATM

Answers are attached above. I need to know the steps to solving this question.

Answer 1

We know that for a first order reaction, t%2 = In2/k At 25°C, t 12 = 377 s k1 = In2/ty = In2/377 s = 0.001839 5-1 At 35°C, t 12 = 333 s k2 = In2/ty = In2/333 s = 0.002082 5-1

We know that the rate of a reaction and the temperature are related by Arrhenius equation as k = Ae RT Let the ky, k, are the rate constants at different temperatures T, T, respectively. Then the Arrhenius equation can be re-write as k = Ae RT In k, = In 4 - Feb In ka = In 4 - Ink - Ink, ha =%267 Given that T, = 25+273.15 = 298.15K T, = 35+273.15 = 308.15K ky _ 0.001839 – 0.883289 k, 0.002082 6. 3.)-(195.15K 305.15K) (i 11 T, (298.15K =(0.003354 -0.003245)K? =(0.000109)K- (2 308.15K In = ln(0.883289) = -0.1241 E -0.1241=- 8.314 J.mol-'K v-1(0.000109) K- 0.1241x8.314 J.mol-? BE=- -=9479.561 J.mol-1 0.000109

Using this Ea value, lets find the rate constant at 50°C

1 - 1 Given that E = 9479.561 J.mol-1 Ty = 25 °C = 25 + 273.15 = 298.15 K Ty = 50°C = 50 + 273.15 = 323.15 K 1 - G 9479.561 J.mol-1 1 8.314 J.mol-K™ 298.15K = -9413.502(0.000259) 323.15K) 8.314 = m () --0295855 ki = -0.295855 = 0.743895 kz=1 ki 0.001839s-1 -=0.002472.5-1 0.743895 0.743895 ką = 0.0024725-1 The rate constant for a first order reaction is

k="ln[4], t [A] where (Alo = initial concentration of A [A] = conc. of A after timet k = rate constant Given that Let initial amount of A = 0.777 M. Amount of A remaining = 0.222 M k = 0.002472 5-1 tu-n [4]. k [A] 002472 -1 In 0.777 0.222 -TX1.252763 0.002472 s-1“ t=506.8699 S = 507 s time required=507 s

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(b) The balanced equation for the given reaction is N2O ---------> N2 + 7 02 Initial After time t 10.00 10.00 - X 0 X 0 2x At half life, 10 - x = half of the initial conc. = 10/2 = 5 That is 10- x= 5 x = 5 Total pressure = (10 -x) + x + 2 x = 10 + 2 x = 10 + 22 (5) = 10 + 2.5 = 12.50 atm Total pressure = 12.50 atm