If `K_(a_1),K_(a_2)` and `K_(a_3)` be the first, second and third dissociation constant of `H_(3)PO_(4)` and `K_(a_1)gtgt K_(2_a) gtgtK_(a_3)` whis is/are correct :
k_(1) and k_(2) are the velocity constants of forward and backward reactions. The equilibrium constant K of the reaction is
For a reaction A overset (k) rarr product, the reaction occurs as : A overset (k_1) rarr A^(**) overset (k_2) rarr B^(**) overset (k_3) rarr product The overall rate constant k is given as: k= 2k_1 (k_2/k_3)^(2//3) If E_(a1) , E_(a2) and E_(a3) are the activation energies of these steps, then what is the relation between overall activation energy E_(a) and the activation energies of these steps.
Find the terms indicated in the following case : a_n =(n-1)(2-n) (3+n), a_1, a_2, a_3 .
If A,A_1,A_2 and A_3 are the areas of the inscribed and escribed circles of a triangle, prove that 1/sqrtA=1/sqrt(A_1)+1/sqrt(A_2)+1/sqrt(A_3)
Why is K_(a_2) << k_(a_1) for H_2SO_4 in water?
A reaction takes place in various steps. The rate constatn for first, second, third and fifth steps are k_(1),k_(2),k_(3) and k_(5) respectively The overall rate constant is given by k=(k_(2))/(k_(3))(k_(1)/(k_(5)))^(1//2) If activation energy are 40, 60, 50, and 10 kJ/mol respectively, the overall energy of activation (kJ/mol) is :
K_(sp) of Zr_(3)(PO_(4))_(4) in terms of solubility (S) is :
For a complex reaction A overset(k) rarr products E_(a1)=180kJ //mol e,E_(a2)=80kJ//mol ,E_(a3)=50kJ//mol Overall rate constant k is related to individual rate constant by the equation k=((k_(1)k_(2))/(k_(3)))^(2//3) . Activation energy ( kJ//mol ) for the overall reaction is : a) 100 b) 43.44 c) 150 d) 140
For a reaction taking place in three steps, the rate constants are k_1, k_2 and k_3 . The overall rate constant k= (k_1 k_2)/k_3 . If the energy of activation values for the first, second and third stages are respectively 40, 50 and 60 kJ mol^-1 , then the overall energy of activation in kJ mol^-1 is
For reaction ArarrB, the rate constant k_(1)=A_(1) e^(-Ea_(1//(RT))) and for the reactio XrarrY, the rate constant k_(2)=A_(2) e^(-Ea_(2//(RT))) . If A_(1)=10^(8),A_(2)=10^(10) and E_(a_(1))=600 cal /mol, E_(a_(2))=1800 cal/mol then the temperature at which k_(1)=k_(2) is (Given :R=2 cal/K-mol) a) 1200 K b) 1200xx4.606 K c) (1200)/(4.606)K d) (600)/(4.606)K
NARENDRA AWASTHI-IONIC EEQUILIBRIUM-Assertin-Reason Type Questions
