Chemical Kinetics and Nuclear ChemistryHard

Question

For hypothetical reaction A → B takes place according to

$$A\overset{\quad K_{1}\quad}{\leftleftarrows}C\left( \text{fast} \right) $$$A + C\overset{\quad K_{2}\quad}{\rightarrow}B\left( \text{slow} \right)$

The rate of reaction is (K1 is equilibrium constant)

Options

A.K2[B][C]
B.K1K2[A]
C.K1K2[A]2
D.K1[B][C]

Solution

$r = K_{2}\lbrack A\rbrack\lbrack C\rbrack\text{ and }K_{1} = \frac{\lbrack C\rbrack}{\lbrack A\rbrack}$

$\therefore r = K_{1}K_{2}\lbrack A\rbrack^{2}$

Create a free account to view solution

View Solution Free
Topic: Chemical Kinetics and Nuclear Chemistry·Practice all Chemical Kinetics and Nuclear Chemistry questions

More Chemical Kinetics and Nuclear Chemistry Questions

For a first order reaction, to obtain a positive slope we need to plot where [A] is theconcentration of reaction of reac...In the reaction : P + Q →R + S the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. ...Activation energy of a chemical reaction can be determined by :...6C12 and 1T3 are formed in nature due to the nuclear reaction of neutron with...The decomposition of H2O2 can be followed by titration with KMnO4 and is found to be a first order reaction. The rate co...