You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

First Order Reaction

If a I-order ...

Question

If a I-order reaction is completed to the extent of $60 %$ and $20 %$ in time intervals, $t_{1}$ and $t_{2}$ what is the ratio, $t_{1} : t_{2}$ ?

6.32

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.587

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.11

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

8.33

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D 4.11
In 1st Order Reaction,

$\frac{t_{1}}{t_{2}} = \frac{l o g (\frac{100}{100 - 60})}{l o g (\frac{100}{100 - 20})}$

$\frac{t_{1}}{t_{2}} = 4.11$

Suggest Corrections

Similar questions

Q. If a first order reaction is completed to the extent of 75% and 50% in time intervals,

t_{1}

and

t_{2}

, what is the ratio,

t_{1} : t_{2}

Q. If a first order reaction is completed to the extent of

75

% and

50

% in time interval,

t_{1}

and

t_{2}

, what is the ratio

t_{1} : t_{2}

Q. Two first order reactions have half-lives in the ratio 8 :1, calculate the ratio of time intervals

T_{1} : T_{2}

. The time

T_{1}

and

T_{2}

are the time period for

{(\frac{1}{4})}^{t h}

and

{(\frac{3}{4})}^{t h}

completion respectively:

Q. Two first order reaction have half-lives in the ratio

8 : 1

. Calculate the ratio of time intervals

t_{1}

and

t_{2}

are the time period for the

{(\frac{1}{4})}^{t h}

and

{(\frac{3}{4})}^{t h}

completion.

The half life of a radio active substance is $20$ days. If $\frac{2}{3}$ part of the substance has decayed in time $t_{2}$ and $\frac{1}{3}$ part of it has decayed in time $t_{1}$ then the time interval between $t_{2}$ and $t_{1} i s (t_{2} - t_{1}) =$

Join BYJU'S Learning Program

If a I-order reaction is completed to the extent of 60% and 20% in time intervals, t1 and t2 what is the ratio, t1:t2?

The correct option is D 4.11In 1st Order Reaction, t1t2=log(100100−60)log(100100−20)t1t2=4.11

If a I-order reaction is completed to the extent of $60 %$ and $20 %$ in time intervals, $t_{1}$ and $t_{2}$ what is the ratio, $t_{1} : t_{2}$ ?

The correct option is D 4.11
In 1st Order Reaction,

$\frac{t_{1}}{t_{2}} = \frac{l o g (\frac{100}{100 - 60})}{l o g (\frac{100}{100 - 20})}$

$\frac{t_{1}}{t_{2}} = 4.11$