What is the rate constant k for thefirst-orderdecomposition of#N_2O_(5(g))#at#25^@C#if the half-life of#N_2O_(5(g))#at that temperature is#4.03 * 10^4#seconds? Under these conditions, what percentage of#N_2O_5#molecules will not have reacted after 1 day?
We know that half-life represents the time needed for a reactant's concentration to reach half of its initial value. The equation for the half-life of a reactant in a first-order reaction is given by
#t_("1/2") = 0.6931/k#, where #k# represents the rate constant.
Since we know that #N_2O_5# has a half-life of #4.03 * 10^4s# under these conditions, the rate constant will be
Since this is a first-order reaction, the equation we'll use to determine what percentage is left unreacted is
#ln(([N_2O_5])/([N_2O_5]_0)) = -k*t#, where
#[N_2O_5]# - the amount that does not react; #[N_2O_5]_0# - the initial amount; #t# - the time of the reaction - in this case, 1 day;
Since #[N_2O_5]# represents the amount that does not react, we can say that #([N_2O_5])/([N_2O_5]_0) * 100%# will be the percentage of #N_2O_5# that does not react.