Advertisements
Advertisements
प्रश्न
The conversion of molecules A to B follow second order kinetics. If concentration of A is increased to three times, how will it affect the rate of formation of B?
Advertisements
उत्तर
For the reaction A → B,
Since it follows second-order kinetics so,
Rate of reaction (r) = k[A]2 ...(1)
If the concentration of reactant increased to three times.
Rate of reaction (r') = k[3A]2 ...(2)
Thus, on dividing equations (1) and (2)
`r/r^' = (k[A]^2)/(k[3A]^2)`
= `1/9`
Therefore, the rate of formation of B increases to 9 times.
APPEARS IN
संबंधित प्रश्न
A → B is a first order reaction with rate 6.6 × 10-5m-s-1. When [A] is 0.6m, rate constant of the reaction is
- 1.1 × 10-5s-1
- 1.1 × 10-4s-1
- 9 × 10-5s-1
- 9 × 10-4s-1
What is pseudo first order reaction? Give one· example of it.
For a reaction, \[\ce{A + B -> Product}\]; the rate law is given by, `r = k[A]^(1/2)[B]^2`. What is the order of the reaction?
How does calcination differ from roasting?
Rate of reaction for the combustion of propane is equal to:
\[\ce{C3H8_{(g)} + 5O2_{(g)} -> 3CO2_{(g)} + 4H2O_{(g)}}\]
For a complex reaction:
(i) order of overall reaction is same as molecularity of the slowest step.
(ii) order of overall reaction is less than the molecularity of the slowest step.
(iii) order of overall reaction is greater than molecularity of the slowest step.
(iv) molecularity of the slowest step is never zero or non interger.
Why does the rate of any reaction generally decreases during the course of the reaction?
Why can’t molecularity of any reaction be equal to zero?
The role of a catalyst is to change
If the 0.05 molar solution of m+ is replaced by a 0.0025 molar m+ solution, then the magnitude of the cell potential would be
