What is the order for the reaction #2"N"_2"O"_5(g) -> 4"NO"_2(g) + "O"_2(g)#?

  1. What is the order of reaction?
  2. What is the concentration of #"O"_2# after #10# minutes?
  3. What is #(Delta["NO"_2])/(Deltat)#?
  4. What is the half-life of #"N"_2"O"_5#?
  5. What is the concentration of #"N"_2"O"_5# after #100# minutes?

2 Answers
Jun 11, 2017

Yes, graphing is the easiest way to do this (or this could be visually challenging!). In fact, learning how to use Excel is a very important skill. I've put a relatively short video on how to use Excel for chemistry here (which goes into graphs near the end):


#1)#

If you JUST put time in minutes on the #x# axis and the concentration of #"N"_2"O"_5# in #"M"# on the #y# axis in Excel, it would turn out to be not linear.

But with some trial and error...

  • graphing the second order integrated rate law does not work; while it does give a positive slope, the graph is curved (not linear), and thus is of the wrong order.

  • graphing the first order integrated rate law works to give a linear graph, confirming it is first order:

I've included the best fit line equation, which is given by:

#bb(underbrace(overbrace(ln["N"_2"O"_5])^(y))_"ln of current conc." = underbrace(overbrace(-k)^(m))_"rate constant"overbrace(t)^(x) + underbrace(overbrace(ln["N"_2"O"_5]_0)^(b))_"ln of initial conc.")#

where #k# is the rate constant, #["N"_2"O"_5]# is the concentration of #"N"_2"O"_5# in #"M"#, and #[" "]_0# means initial concentration. You know that #t# means time in #"min"#.

From looking at the best fit line equation, here is what you can immediately get from it:

#"slope" = -k#

#=> k = -"slope"#

#=> color(blue)(k = "0.0301 min"^(-1))#

#"y-int." = -4.3894#

#=> color(blue)(["N"_2"O"_5]_0) = e^("y-int.") = e^(-4.3894) = color(blue)("0.0124 M")#

And furthermore, the rate law for this would therefore be (from knowing what the order is):

#r(t) = k["N"_2"O"_5]#

#2)# At #t = "10 min"#, to get the concentration of #"O"_2#, we can examine the concentration of #"N"_2"O"_5# remaining, #"0.0092 M"#, and determine the amount of #"O"_2# currently made.

This may or may not be a new concept, but let's consider something called an ICE Table (initial, change, equilibrium), which is normally a way to track the changes in concentration in a reaction on its way to "equilibrium", the point when the reaction has the same rate forwards and backwards.

In the ICE Table below, the coefficients in front of #x# correspond to the coefficients in the chemical reaction.

#2"N"_2"O"_5(g) -> 4"NO"_2(g) + "O"_2(g)#

#"I"" ""0.0124 M"" "" ""0 M"" "" "" ""0 M"#
#"C"" "-2x" "" "" "+4x" "" "" "+x#
#"E"" ""0.0092 M"" "" "4x" "" "" "" "x#

Well, we can track the progress of the reaction using the reaction quotient, #Q#, even though it hasn't reached equilibrium yet. It's the same kind of definition, but we are using the current concentrations instead.

Since we know that the change in concentration of #"N"_2"O"_5# was #2x = "0.0032 M"#, we know that #color(blue)(x = "0.0016 M")# is the change in concentration of #"O"_2#.

And since #"O"_2# started at #"0 M"# (i.e. the reaction started with only reactant), that is its concentration at #bb"10 min"#!

#3)# The initial rate of production of #"NO"_2# comes from the rate law, since we know the rate constant and the starting concentration of #"N"_2"O"_5#:

#r(t) = k["N"_2"O"_5]#

#r(t) = 3.735 xx 10^(-4) "M/min" = |overbrace(-)^"consumption"1/2 overbrace((Delta["N"_2"O"_5])/(Deltat))^"Initial rate"|#

#= overbrace(+)^"production"1/4 overbrace((Delta["NO"_2])/(Deltat))^"Initial rate"#

So, it is:

#color(blue)((Delta["NO"_2])/(Deltat)) = 4/2 xx 3.735 xx 10^(-4) "M/min"#

#= color(blue)(7.47 xx 10^(-4) "M/min")#

#4)# The half-life is known to be at the point where #["N"_2"O"_5] = 1/2["N"_2"O"_5]_0#, i.e. when the reactant concentration has halved. We would obtain this from the integrated rate law:

#ln["N"_2"O"_5] = -kt + ln["N"_2"O"_5]_0#

First, plug in #["N"_2"O"_5] = 1/2["N"_2"O"_5]_0#:

#ln(1/2["N"_2"O"_5]_0) = -kt_"1/2" + ln["N"_2"O"_5]_0#

Now use the property that #lna - lnb = ln(a/b)#:

#ln\frac(1/2cancel(["N"_2"O"_5]_0))(cancel(["N"_2"O"_5]_0)) = -kt_"1/2"#

#=> -ln2 = -kt_"1/2"#

Thus, the first-order half-life is given by:

#=> color(green)(t_"1/2" = (ln2)/k)#

And now, we can get:

#color(blue)(t_"1/2") = (ln2)/("0.0301 min"^(-1))#

#=# #color(blue)("23.03 min")#

#5)# At #"100 min"#, then, we should expect that

#"100 mins passed"/"23.03 min half-life" ~~ 4.34# half-lives passed.

Therefore, the initial concentration #["N"_2"O"_5]_0# should have halved #4.34# times, or been multiplied by #1/2 cdots 1/2 = (1/2)^(4.34) = 1/(2^(4.34))#.

This means the concentration after #"100 min"# is approximately:

#color(blue)(["N"_2"O"_5]_("100 min.")) = "0.0124 M" xx 1/(2^(4.34)) ~~ color(blue)(6.11 xx 10^(-4) "M")#

(or #0.0611 xx 10^(-2) "M"#.)

Jun 11, 2017

Answer:

Kinetic Rate Law for Decomposition of #N_2O_5#
#"Rate" = 0.100 min^-1 [N_2O_5]^1#

Explanation:

Kinetics of #N_2O_5# Decomposition:
enter image source here