Chemical Kinetics — Page 9 | JEE Chemistry

Q81

The integrated rate equation is Rt = log C0 - log Ct . The straight line graph is obtained by plotting

A time vs log Ct

B

{1 \over {time}}

vs Ct

C time vs Ct

D

{1 \over {time}}

vs

{1 \over {{C_t}}}

Correct Answer

Option A

Solution

Rt = \log {C_o} - {{\mathop{\rm logC}\nolimits} _t}

It is clear from the equation that if we plot a graph between

\log \,{C_t}

and time, a straight line with a slope equal to

- {k \over {2.303}}

and intercept equal to

\log \,\left[ {{A_o}} \right]

will be obtained.

Q82

N2O5 decomposes to NO2 and O2 and follows first order kinetics. After 50 minutes, the pressure inside the vessel increases from 50 mmHg to 87.5 mmHg. The pressure of the gaseous mixture after 100 minute at constant temperature will be :

A 175.0 mmHg

B 116.25 mmHg

C 136.25 mmHg

D 106.25 mmHg

Correct Answer

Option D

Solution

N2O5 $\rightleftharpoons$ 2NO2 +

{1 \over 2}

O2 At t = 0 50 0 0 At t = 50 min 50 $-$ P1 2P1

{{{P_1}} \over 2}

Total pressure after 50 min, = 50 $-$ P1 + 2P1 +

{{{P_1}} \over 2}

= 87.5

\therefore\,\,\,

50 +

{{3{P_1}} \over 2}

= 87.5 $\Rightarrow$

\,\,\,

P1 = 25 So, 50 min is the Half $-$ life period of the reaction.

Then 100 min is two half life .tg .tg N2O5 $\rightleftharpoons$ 2NO2 +

{1 \over 2}

O2 At t = 100 min 50 $-$ P2 2P2

{{{P_2}} \over 2}

\therefore\,\,\,

50 $-$ P2 =

{{25} \over 2}

P2 = 37.5

\therefore\,\,\,

Total pressure at t = 100 min = 50 $-$ P2 + 2P2 +

{{{P_2}} \over 2}

= 50 +

{3 \over 2}

$\times$ 37.5 = 106.25 mm of Hg

Q83

For a first order reaction, the time required for completion of 90% reaction is 'x' times the half life of the reaction. The value of 'x' is (Given : ln 10 = 2.303 and log 2 = 0.3010)

A 1.12

B 2.43

C 3.32

D 33.31

Correct Answer

Option C

Solution

\mathrm{A} \rightarrow

Products For a first order reaction,

\mathrm{t}_{1 / 2}=\frac{\ln 2}{\mathrm{k}}=\frac{0.693}{\mathrm{k}}

Time for

90 \%

conversion,

t_{90 \%}=\frac{1}{k} \ln \frac{100}{10}=\frac{\ln 10}{k}=\frac{2.303}{k}

t_{90\%}=\frac{2.303}{0.693} t_{1 / 2}=3.32 t_{1 / 2}

Q84

In the following reaction; xA

\to

yB

{\log _{10}}\left[ { - {{d\left[ A \right]} \over {dt}}} \right] = {\log _{10}}\left[ {{{d\left[ B \right]} \over {dt}}} \right] + 0.3010

'A' and 'B' respectively can be :

A n-Butane and Iso-butane

B C2H4 and C4H8

C C2H4 and C6H6

D N2O4 and NO2

Correct Answer

Option B

Solution

xA $\to$ yB

{1 \over x}\left\{ { - {{d\left[ A \right]} \over {dt}}} \right\} = {1 \over y}\left\{ {{{d\left[ B \right]} \over {dt}}} \right\}

$\Rightarrow$

- {{d\left[ A \right]} \over {dt}} = {x \over y}\left\{ {{{d\left[ B \right]} \over {dt}}} \right\}

Taking log both sides, we get $\Rightarrow$

{\log _{10}}\left[ { - {{d\left[ A \right]} \over {dt}}} \right] = {\log _{10}}\left[ {{{d\left[ B \right]} \over {dt}}} \right] + {\log _{10}}\left( {{x \over y}} \right)

Given that,

{\log _{10}}\left[ { - {{d\left[ A \right]} \over {dt}}} \right] = {\log _{10}}\left[ {{{d\left[ B \right]} \over {dt}}} \right] + 0.3010

So, by comparing both of them we get

{\log _{10}}\left( {{x \over y}} \right)

= 0.3010 =

{\log _{10}}\left( 2 \right)

$\therefore$

{{x \over y}}

= 2 $\Rightarrow$ x = 2y If x = 2 then y = 1 The reaction is of type 2A $\to$ B So possible reaction is 2C2H4 $\to$ C4H8