What is the order for the reaction #2"N"_2"O"_5(g) > 4"NO"_2(g) + "O"_2(g)#?
 What is the order of reaction?
 What is the concentration of
#"O"_2# after #10# minutes?
 What is
#(Delta["NO"_2])/(Deltat)# ?
 What is the halflife of
#"N"_2"O"_5# ?
 What is the concentration of
#"N"_2"O"_5# after #100# minutes?
 What is the order of reaction?
 What is the concentration of
#"O"_2# after#10# minutes?  What is
#(Delta["NO"_2])/(Deltat)# ?  What is the halflife of
#"N"_2"O"_5# ?  What is the concentration of
#"N"_2"O"_5# after#100# minutes?
2 Answers
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):
If you JUST put time in minutes on the
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))#
#"yint." = 4.3894#
#=> color(blue)(["N"_2"O"_5]_0) = e^("yint.") = 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]#
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
#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,
Since we know that the change in concentration of
And since
#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")#
#ln["N"_2"O"_5] = kt + ln["N"_2"O"_5]_0#
First, plug in
#ln(1/2["N"_2"O"_5]_0) = kt_"1/2" + ln["N"_2"O"_5]_0#
Now use the property that
#ln\frac(1/2cancel(["N"_2"O"_5]_0))(cancel(["N"_2"O"_5]_0)) = kt_"1/2"#
#=> ln2 = kt_"1/2"#
Thus, the firstorder halflife 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")#
#"100 mins passed"/"23.03 min halflife" ~~ 4.34# halflives passed.
Therefore, the initial concentration
This means the concentration after
#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