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The rate of a reaction aA +bB Products, is given by  \mathrm{k[A]^{n} [B]^{m}} . On doubling the concentration of A and doubling the volume of the vessel the rate of reaction changes by 

 

Option: 1

\mathrm{2^{-n} \\ }


Option: 2

2^{(\mathrm{n}+\mathrm{m})} \


Option: 3

2^{(\mathrm{n}-\mathrm{m})} \\


Option: 4

2^{-\mathrm{m}}


Answers (1)

best_answer

By doubling the volume of reaction vessel, the concentration of both the reactants halve. 

\text { rate }_1=k[A]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}} \\
\text { rate }_2=\mathrm{k}[2 \mathrm{~A} / 2]^{\mathrm{n}}[\mathrm{B} / 2]^{\mathrm{m}}

=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}}(1 / 2)^{\mathrm{m}}

=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}} \times 2^{-\mathrm{m}}=\text { rate }_1 \times 2^{-\mathrm{m}}
Hence, (4)

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