The dependence of rate on concentrations of reactants is expressed in terms of rate law…
Chemistry · JEE Advanced · NTA Exams — Chemical Kinetics
The dependence of rate on concentrations of reactants is expressed in terms of rate law, which is established experimentally.
\(\text { Rate }=\mathrm{k}[\mathrm{~A}]^{\mathrm{a}}[\mathrm{~B}]^{\mathrm{b}}\) (Rate law) ..........(i)
The exponents a, b, etc. (determined experimentally) may or may not be equal to the respective stoichiometric coefficients. k is the velocity constant of the reaction. The determination of rate law is simplified by the isolation method in which the concentration of all the reactants except one are in large excess. If B is in large excess. we can approximate [B]by \([\mathrm{B}]_{0}\)
Hence, \(\text { Rate }=\mathrm{k}[\mathrm{~A}]^{\mathrm{a}}[\mathrm{~B}]^{\mathrm{b}}=\mathrm{k}[\mathrm{~A}]^{\mathrm{a}}[\mathrm{~B}]_{\mathrm{b}}^{\mathrm{b}}=\mathrm{k}^{\prime}[\mathrm{A}]^{\mathrm{a}}\) \(\left(\mathrm{k}^{\prime}=\mathrm{k}[\mathrm{~B}]_{0}^{\mathrm{b}}\right)\)
or \(\log (\text { initialrate })=\log\) \(\mathrm{v}_{0}=\log \mathrm{k}^{\prime}+\mathrm{a} \log [\mathrm{~A}]\) ...... (ii)
A plot of log (rate) against \(\log [\mathrm{A}]\) values will be a straight line which enables to calculate both \(\mathrm{k}^{\prime}\) and a. Similarly orders with respect to other reactants. taken in much smaller concentrations turn by turn, can be determined. Consider the reaction : \(2 \mathrm{I}_{(\mathrm{g})}+\mathrm{A}_{(\mathrm{g})} \longrightarrow \mathrm{I}_{2(\mathrm{~g})}+\mathrm{A}_{(\mathrm{g})}\)
The following figures show the variation of log υ0 against (a ) \(\log \mathrm{I}_{0}\) for a given \([\mathrm{Ar}]_{0}\) and ( b) \(\log [\mathrm{Ar}]_{0}\) for a given \([\mathrm{I}]_{0}\)
The rate constants of most reactions increase as the temperature is increased. The rate constant increases by about \(100-200 \%\) for a temperature rise of \(10 \mathrm{~K}\). It is found experimentally for many reactions that a plot of \(\ell \mathrm{nk}\) against \(1 / T\) gives a straight line. This behaviour is expressed in the form of equation.
\(\ell \mathrm{n} \mathrm{k}=\ell \mathrm{n} \mathrm{~A}-\frac{\mathrm{E}_{\mathrm{a}}}{\mathbb{R}}\) .............. (iii)
The rate of change of molar concentration of C in Reaction – 1, is found to be 3.0 × 10–3 mol L–1s–1. The rate of reaction and rate of disappearance of the reactant B are respectively.
- 3.0 × 10–3 mol L–1s–1 each
- 1.0 × 10–3 mol L–1s–1 each
- 1.0 × 10–3 mol L–1s–1 and 2.0 × 10–3 mol L–1 s–1
- None of these.
Answer
(C) 1.0 × 10 –3 mol L –1 s –1 and 2.0 × 10 –3 mol L –1 s –1
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