CBSE Class 12 Chemistry 2015 Solved Paper

Question : 24 of 26

Marks: +1, -0

(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(Given

\log 2=0.3010, \log 4=0.6021

)
OR
(a) For a reaction

{\A}+ {\B} → {\P}

, the rate is given by Rate

=k[ {\ A }][ {\B}]{^2}

(i) How is the rate of reaction affected if the concentration of

B

is doubled ?
(ii) What is the overall order of reaction if

A

is present in large excess?
(b) A first order reaction takes

{3 0}

minutes for

{5 0} \%

completion. Calculate the time required for

{9 0 \%}

completion of this reaction.

(\log 2=0.3010)

Solution:

\;k=\;{2.303}/{t} \log \;{ [ {\A}_0] }/{[ {\A}]}

\;k=\;{2.303}/{30} \log \;{0.60}/{0.30}

\;k=\;{2.303}/{30} × 0.301=0.023 {\ s}^{-1}

\;k=\;{2.303}/{60} \log \;{0.60}/{0.15}

\;k=\;{2.303}/{60} × 0.6021=0.023 {\ s}^{-1}

(ii) As

k

is constant in both the readings, hence it is a pseudo-first order reaction.

\;\text " Rate "\;\;=-\;{∆[ {\R}]}/{∆ t}

\;=-\;{[0.15-0.30]}/{60-30}

\;=0.005 {\mol} {\L}^{-1} {\ s}^{-1}

OR
(a) (i) Rate will increase 4 times of the actual rate of reaction.
(ii) Second order reaction
(b)

t_{1 \/ 2}\;=\;{0.693}/{k}

30 {\min}\;=\;{0.693}/{k}

k\;=0.0231 {\min}^{-1}

t\;=\;{2.303}/{t} \log \;{ [ {\A}_0] }/{[ {\A}]}

t\;=\;{2.303}/{0.0231} {\min}

t\;=99.7 {\min}

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