The reaction 2N(2)O(5)hArr 2N(2)O(4)+O(2...

The reaction `2N_(2)O_(5)hArr 2N_(2)O_(4)+O_(2)` is

Bimolecular and second order

Unimolecular and first order

Bimolecular and first order

Bimolecular and zero order

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The correct Answer is:

This reaction is bimolecular and first order reaction

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The reaction 2N_(2)O_(5) hArr 2NO_(2)+O_(2) follows first order kinetics. Hence, the molecularity of the reaction is

The half life for the reaction N_(2)O_(5) hArr 2NO_(2)+1/2 O_(2) in 24 hr at 30^(@)C . Starting with 10 g of N_(2)O_(5) how many grams of N_(2)O_(5) will remain after a period of 96 hours ?

For the reaction 2NO_(2)(g)+(1)/(2)O_(2)(g)hArrN_(2)O_(5)(g) if the equilibrium constant is K_(p) , then the equilibrium constant for the reaction 2N_(2)O_(5)(g)hArr4NO_(2)(g)+O_(2)(g) would be :

The reaction 2N_(2)O_(5)(g)rarr4NO_(2)(g)+O_(2)(g) is found to be firt order with respect to N_(2)O_(5)(g) . Which of the following is correct ?

For the reaction : 2N_(2)O_(5) rarr 4NO_(2) +O_(2) , the rate of reaction in terms of O_(2) is d[O_(2)] /dt. In terms of N_(2)O_(5) , it will be :

For the reaction 2NO_(2(g))+(1)/(2)O_(2)hArrN_(2)O_(5(g)) if the equilibrium constant is K_(P) ,then the equilibrium constant for the reaction 2N_(2)O_(5(g))hArr4NO_(2(g))+O_(2(g)) would be :-

The reaction N_(2)O_(5) to 2NO_(2)+I//2O_(2) is of first order in N_(2)O_(5) . Its rate constant is 6.2 xx 10^(-6)s^(-1) . If the beginning N_(2)O_(5) is 15 mol L^(-1) , calculate the rate of reaction in the beginning.

If in the reaction, N_(2)O_(4)(g)hArr2NO_(2)(g), alpha is the degree of dissociation of N_(2)O_(4) , then the number of moles at equilibrium will be

ERRORLESS-CHEMICAL KINETICS-NCERT BASED QUESTIONS (Rate Law and Rate Constant)

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