Question

The dissociation of nitrogen pentoxideis a first order reaction. In first 24 min. 75% of nitrogen pentoxide is dissociated. What amount of nitrogen pentoxidewill be left behind after one hour of start of reaction?

Answer 1

Dear ​Hardee Shah,

$$\begin{gathered} For1storderreaction \\ k=\frac{2.303}{t}log\frac{a}{a-x} \\ where;k=rateconstant \\ t=time(inmin) \\ a=initialconc. \\ a-x=finalconc. \\ k=\frac{2.303}{24}log\frac{1}{0.25}=0.058mi{n}^{-1} \\ soafter1hror60minamountof{N}_2{O}_{5}leftis(a-x) \\ 0.058=\frac{2.303}{60}log\frac{1}{a-x} \\ log\frac{1}{a-x}=1.51 \\ \frac{1}{a-x}={10}^{1.51}=32.36 \\ a-x=\frac{1}{32.36}=0.031 \\ or3.1\%of{N}_2{O}_{5}willbeleftafter60min. \\ Regards \end{gathered}$$