Question

The reaction HCN (aq) + 2H2O (l) > NH4HCO2 (aq)   is first order, and it's rate =k [HCN]. The rate constant k at 65°C is 8.06 ×10 -8 s-1. How long will it take for thr concentration of the HCN solution to drop from an initial 0.0800M to 0.0600M at this temperature? What is the half life of thr reaction

Answer 1

k = 8.06 ×10 -8 s-1

For a first order reaction,

ln[A]t/[Ao]=-kt where [A]t=concentration of reactant at time t

                               [Ao]=initial conc of reactant

ln (0.0600M)/(0.0800M)=-8.06 ×10 -8 s-1*t

ln 0.75=-8.06 ×10 -8 s-1*t

2.303 log (0.75)=-8.06 ×10 -8 s-1*t

t=2.303 log (0.75/-8.06 ×10 -8 s-1=2.303*(-0.1249)/-8.06 ×10 -8 s-1=0.03568*10^8 s=3.568 * 10^6 s

At t1/2=half life, [A]t=1/2 [A]o

so ln[A]t/[A]o==-kt

ln 1/2=-kt1/2

or 0.693/K=t1/2

so t1/2=0.693/8.06 ×10 -8 s-1=0.08598*10^8 s=8.598 *10^6 S