Question

12. A zero order reaction is 50% complete in 20 minutes. What percentage would be completed at the end of 30 minutes? In how many minutes would the concentration be reduced to zero?​

Answer 1

Answer:

20%completedreactionmeans

leftamountA=80

initialamountA

∘

=100andthereactioniscompletedint=10s

Similarlyfor50%,

leftamountA=50

initialamountA

∘

=100

A=A

∘

−kt

80=100−k×10

k×10=20

k=2moll

−1

s

−1

A=A

∘

−kt

50=100−2×t

2t=50

t=25s

Answer 2

Answer:

In the 30 minutes, 75% of reaction is completed.

The concentration of reactants would be reduced to zero in 40 minutes.

Explanation:

Consider that initial concentration of reactant = $A_{o}$

Given $t_{\frac{1}{2} =20 min$

For zero order reaction: $t_{\frac{1}{2} }=\frac{A_{o}}{2k}$

⇒   $k=\frac{A_{o}}{2 t_{\frac{1}{2} }}=\frac{A_{o}}{40 }}$

• When t=30min, $A_t$ conc. of reactants at time t

         For zero order reaction:

         $A_{t}=A_o-kt\\A_t=A_o-\frac{A_o}{40}(30)\\A_t=A_o-\frac{A_o}{4}(3)\\A_t=\frac{4A_o-3A_o}{4}\\ A_t=\frac{A_o}{4}$

         Reacted conc. of reactants at time t $=A_o-A_t=A_{o}-\frac{A_{o}}{4}=\frac{3A_{o}}{4}$

         Percentage of reaction completed $=\frac{\frac{3A_{o}}{4} }{A_{o}} *100=\frac{3}{4}*100$  = 75%

• When 100% reaction is completed.

        Time required to complete reaction  $=2t_{\frac{1}{2} }=2(20)=40 min$