Consider the decomposition of N 2 O 5 as shown in the following equation to answ
ID: 873912 • Letter: C
Question
Consider the decomposition of N2O5 as shown in the following equation to answer the next three questions:
2 N2O5(g) --> 4 NO2(g) + O2(g)
1.If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of NO2 and O2 respectively?
2.The reaction shown above has been determined to be first order with respect to N2O5. If 0.0250 moles of N2O5 are charged into a 2.00 L vessel, how many moles will remain after 150 seconds if the rate constant is 6.82 x 10-3 s-1 (at 70 oC)?
3.What is the half-life for this reaction at 70 oC?
Explanation / Answer
2 N2O5 (g) --> 4 NO2 (g) + O2 (g)
d[N2O5]/dt = 4.2X10-7M/s
1. rate of reaction, r = - 1/2 d[N2O5]/dt = 1/4 d[NO2]/dt = d[O2]/dt
d[NO2]/dt = 4/2 d[N2O5]/dt
= 2* 4.2X10-7M/s
= 8.4X10-7M/s
d[O2]/dt = 1/2 d[N2O5]/dt
= 1/2* 4.2X10-7M/s
= 2.1X10-7M/s
2. Initial concentration of [N2O5]o = 0.0250/ 2
= 0.0125 M
We have, ln[A] = ?kt + ln[A]o
ln[N2O5] = - kt + ln[N2O5]o
k = 6.82X10-3 /s & t = 150s
Substituting these values in above equation,
ln[N2O5] = - (6.82X10-3)*150 + ln(0.0125)
= - 1.023 - 4.382
= - 5.405
[N2O5] = e - 5.405
[N2O5] = 4.494X10-3 M
3. t1/2 = 0.693/ k
= 0.693/ 6.82X10-3 s
= 101.61 s