A first order reaction has a specific reaction rate of 10 -2 sec -1 . How much time will it take for 20 g of the reactant to reduce to 5 g ?

Reaction rate = 10 - 2 sec - 1 For a first order reaction ⇒ K = 1 t ln A 0 A t Here, we can use the weight of the reactant instead of concentration. Or , 10 - 2 sec - 1 = 1 t ln 20 5 Or , t = 1 10 - 2 sec - 1 ln 4 Or , t = 1 . 386 10 - 2 sec - 1 = 138 . 6 sec

What will be the effect of temperature on rate constant?

The rate constant of a reaction is doubled or tripled with the 10 o C or K rise in temperature. However, the exact relation between rate of a chemical reaction and temperature is given by the Arrhenius equation, k = A e - E a R T Where, k = rate constant A = Arrhenius factor or the frequency factor T = temperature R = gas constant E a = activation energy Thus, the rate constant of a chemical reaction normally increases with increase in temperature.

For a reaction: 2 NH 3 ( g ) ⟶ Pt N 2 ( g ) + 3 H 2 ( g ) : Rate = k . Write the unit of k .

The given reaction, 2 NH 3 ( g ) ⟶ Pt N 2 ( g ) + 3 H 2 ( g ) : The rate expression is, Rate = k , as clearly evident from the rate expression, that no species is involved in the rate law, that concludes the zero order nature of the reaction. Accordingly, Rate = k mol L - s - 1 = k Therefore, the unit of k is mol L - s - 1

For a reaction: 2 NH 3 ( g ) ⟶ Pt N 2 ( g ) + 3 H 2 ( g ) : Rate = k . Write the order and molecularity of this reaction.

The given reaction, 2 NH 3 ( g ) ⟶ Pt N 2 ( g ) + 3 H 2 ( g ) : The rate expression is, Rate = k , as clearly evident from the rate expression, that no species is involved in the rate law, that concludes the zero order nature of the reaction, and on contrast the molecularity is two as two molecules are participating in the reaction.

For a reaction A + B → P , the rate is given by : Rate = k A B 2 . What is the overall order of reaction if A is present in excess ?

Rate law is given as : Rate = k A B 2 . Reactant present in excess does not affect the order of reaction and hence, the rate of reaction is independent of the concentration of such reactants. Hence,when A is present in excess, rate can be expressed as : Rate = k B 2 . Hence, the overall order of reaction is two.

For a reaction A + B → P , the rate is given by : Rate = k A B 2 . How is rate of reaction affected if the concentration of B is doubled ?

Rate law for the reaction is given as : Rate = k A B 2 . When the concentration of reactant B is doubled, rate can be expressed as : Rate = k A 2 B 2 = 4 k A B 2 . Hence, when the concentration of reactant B is doubled, the rate of reaction becomes four times or its initial value.

For a zero order reaction will the molecularity be equal to zero? Explain.

The molecularity of a reaction is equal to the number of molecules participating in the elementary reaction. It is equal to the number of stoichiometric coefficients present in the balanced chemical reaction. The order of the reaction is experimentally determined. It is equal to the molecularity of the reaction for the elementary reaction, except for a zero-order. No, the molecularity is an aspect that can never be zero or a fractional number as it displays the number of reactants that take part in a reaction, which can never be zero. It cannot be determined for complex reactions.

For a first order reaction, show that the time required for 99% completion is twice the time required for the completion of 90% of reaction.

In this class of reactions, the rate of the reaction is proportional to the first power of the concentration of the reactant R. The first-order rate equation can be written in the form: t = 2 . 303 k log R 0 R t 99 % = 2 . 303 k log 100 1 = 2 . 303 k × 2 t 90 % = 2 . 303 k log 100 10 = 2 . 303 k t 99 % t 90 % = 2 . 303 k × 2 2 . 303 k = 2 t

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter -1 - Chemical Kinetics>Assam Board-2020>Q 11

EASY

12th Assam Board

IMPORTANT

Earn 100

A reaction is first order in A and second order in B.

Write the differential rate equation.

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