Answer to Question #166046 in General Chemistry for dale

Question

The decomposition of NO_2(g) : 2NO_2(g) ----->2NO_(g) +O_(g) is a second order reaction and the rate constant is 0.755L/(mol-s). In an experiment, the initial concentration of NO_2(g) was 0.00650 mol/L. (a)What is the concentration of NO_(g) after 125 seconds have elapsed? (b) How many seconds will it take for the concentration of NO_2(g) to drop to 0.00100 mol/L?

Answer 1

(a) For second order reaction:

$k = \frac{1}{t}[\frac{1}{A_t} -\frac{1}{A_0}] \\ 0.755 = \frac{1}{125}[\frac{1}{A_t} -\frac{1}{0.0065}] \\ 94.375 = \frac{1}{A_t} -\frac{1}{0.0065} \\ 248.221 = \frac{1}{A_t} \\ A_t = 0.00402 \;M$

$(b) \; 0.755 = \frac{1}{t}[\frac{1}{0.00100} -\frac{1}{0.0065}] \\ 0.755 = \frac{1}{t}[1000 – 153.84] \\ 0.755 = \frac{846.16}{t} \\ t = 1120 \;s$