Question #a5353

1 Answer
Dec 6, 2015

Answer:

#(a)#

#5.4xx10^(-3)#

#(b)#

#101"s"#

Explanation:

(a)

This is a 1st order reaction so:

#R=k[N_2O_5]^1#

We can integrate this rate law (I won't go into the maths here) to give:

#N_t=N_0e^(-kt)#

#N_t# is the no. moles after time #t#.

#N_0# is the initial no. of moles.

Taking natural logs of both sides#rArr#

#lnN_t=lnN_0-kt#

Now we put in the values:

#lnN_t=ln(0.015)-(6.82xx10^(-3)xx2.5xx60)#

(note we must convert "min" to "s" so x t by 60)

#:.lnN_t=-5.223#

#:.N_t=0.0054"s"#

(b)

To get the 1/2 life we use:

#t_(1/2)=0.693/k#

Again, I won't go into the derivation of this.

#:.t_(1/2)=(0.693)/(6.82xx10^(-3))=101"s"#