Home
Class 12
CHEMISTRY
The rate law for a reaction is found to ...

The rate law for a reaction is found to be : Rate `=k [NO_(2)^(-) ][ I^(-)][H^(+)]^(2)`
How would the rate of reaction change when (i) Concentration of `H^(+)` is doubled
(ii) Concentration of `I^(-)` is halved (iii) Concentration of each of `NO_(2)^(-),I^(-)` and `H^(+)` are tripled?

Text Solution

AI Generated Solution

To solve the problem, we need to analyze how changes in the concentrations of the reactants affect the rate of the reaction based on the given rate law: **Rate = k [NO2^(-)][I^(-)][H^(+)]^2** Let’s denote: - [NO2^(-)] = A - [I^(-)] = B - [H^(+)] = C ...
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL KINETICS

    PRADEEP|Exercise PROBLEMS FOR PRACTICE|73 Videos

  • CHEMICAL KINETICS

    PRADEEP|Exercise CURIOSITY QUESTION|2 Videos

  • BIOMOLECULES

    PRADEEP|Exercise IMPORTANT QUESTIONS (FOR BOARD EXAMINATION)|25 Videos

  • CHEMISTRY IN EVERYDAY LIFE

    PRADEEP|Exercise IMPORTANT QUESTION FOR BOARD EXAMINATION|30 Videos

Similar Questions

Explore conceptually related problems

The rate law for a certain reaction is found to be : Rate = k[A] [B]^(2) How will the rate of this reaction compare if the concentration of A is doubled and the concentration of B is halved ? The rate will :

The experiment rate law for the reaction S_(2)O_(8)^(2-)(aq) + 2I^(ɵ)(aq) rarr 2SO_(4)^(2-) (aq) + I_(2)(aq) is k[S_(2)O_(8)^(2-)] [I^(ɵ)] . How would the rate change if (a) Concentration of S_(2)O_(8)^(2-) is halved. (b) Concentration of S_(2)O_(8)^(2-) and I^(ɵ) are halved.

The rate law for the following reactions: Ester + H^(o+) rarr "Acid" + "Alcohol" , is dx//dt = k(ester)[H_(3)O^(o+)]^(0) What would be the effect on the rate if (a) concentration of ester is doubled. (b) concentration of H^(o+) ion is doubled.

The rate law for the reaction RCl+NaOH (aq) to ROH +NaCl is given by rate k_(1)[RCl] the rate of reaction will be doubled on doubling the concentration of NaOH.

In the reaction 2A+B to A_2B , Rate = k[A]^2 [ B ] if the concentration of A is doubled and that of B is halved, then the rate of reaction will:

For the reaction A+BrarrC , the rate law is : Rate=k [A]^(2) . Which change(s) will increase the rate of the reaction? (P) Increasing the concentration of A (Q) Increasing the concentration of B