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)
According to the fig-1 (a ) and 1 ( b), the rate law for the reaction
2I(g) + Ar(g)
I2(g) + Ar(g) is given by- Rate = k [I] [Ar]
- Rate = k[I]2
- Rate = k[I]2 [Ar]2
- Rate = k[I]2[Ar]
Answer
(D) Rate = k[I] 2 [Ar]
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- In the reaction, P + Q → R + S , the time taken for 75% reaction of P is twice the time taken for 50%…
- Consider a reaction aG + bH → products. When concentration of both the reactants G and H is doubled, the rate…
- For the reaction A + B → C, it found that doubling the concentration of A increases the rate by 4 times, and…
- The decomposition of ozone is believed to occur by the mechanism : O 3 ⇌ O 2 + O (fast) O + O 3 → O 2 (slow)…
- In the decomposition of Ammonia it was found that at 50 torr pressure T 1/2 was 3.64 hour while at 100 torr T…
- For a first order reaction A → P, the temperature (T) dependent rate constant (k) was found to follow the…
- Four vessels 1, 2, 3 and 4 contain respectively, 10 mol atom (t 1/2 = 10 hours), 1mol atom (t 1/2 = 5 hours)…
- Assertion (A): According to steady state hypothesis, in a multistep reaction, the change in concentration…
More Chemical Kinetics questions · Browse all practice questions