The translational kinetic energy `E=`_____
`(3)/(2)K_(B)NT`
`(2)/(3)K_(B)NT`
`(1)/(3)nmv^(2)`
`(3)/(2)nmv^(2)`

At 25^(@) C the equilibrium constant K,and K_(2) of two reaction are 2NH_(3)hArrN_(2)+3H_(2) ,K_(1) (1)/(2)N_(2)+(3)/(2)H_(2)hArrNH_(3),K_(2) The relation b//w two equilibrium constant is :-

If log_(e )k=2[(3)/(5)+(1)/(3)((3)/(5))^(3)+(1)/(5)((3)/(5))^(5)+….oo] then k =

Which would exhibit ionisation isomerism (a) [Co(NH_(3))_(6)][(C_(2)O_(4))_(3)] (b) [Co(NH_(3))_(5)Br]^(2+)SO_(4)^(2-) (c ) K_(3)[Fe(CN)_(6)] (d) K_(3)[Fe(C_(2)O_(4))_(3)] .

If v=(3hati+2hatj+6k) m/s and m=(2)/(7)kg then find kinetic energy (i.e. (1)/(2)mv^(2))

For the reaction 2AtoB+3C , if -(d[A])/(dt)=k_(1)[A]^(2),-(d[B])/(dt)=k_(2)[A]^(2),-(d[C])/(dt)=k_(3)[A]^(2) the correct reaction between k_(1),k_(2) and k_(3) is :

The total energy of molecules is divided equally amongst the various degrees of freedom of a molecule. The distribution of kinetic energy along x, y, z axis are E_(K_(x)), E_(K_(y)), E_(K_(z)) Total K.e =E_(K_(x)) + E_(K_(y)) + E_(K_(z)) Since the motion of molecule is equally probable in all the three directions, therefore E_(K_(x)) = E_(K_(y)) = E_(K_(z)) =1/3 E_(K) =1/3 xx 3/2 kT = 1/2 kT , where k =R/N_(A) = Botzman constant. K.E. = 1/2 kT per molecule or =1/2 RT per mole. In vibration motion, molecules possess both kinetic energy as well as potential energy. This means energy of vibration involves two degrees of fiuedom. Vibration energy =2 xx 1/2kT =2 xx 1/2RT [ therefore two degrees of freedom per mole] If the gas molecules have n_(1) translational degrees of freedom, n_2 rotational degrees of freedom and n_(3) vibrational degrees of freedom, that total energy = n_(1)[(kT)/2] + n_(2) [(kT)/2] + n_(3) [(kT)/2] xx 2 Where 'n' is atomicity of gas. The vibrational kinetic energy of CO_2 molecule is

The total energy of molecules is divided equally amongst the various degrees of freedom of a molecule. The distribution of kinetic energy along x, y, z axis are E_(K_(x)), E_(K_(y)), E_(K_(z)) Total K.e =E_(K_(x)) + E_(K_(y)) + E_(K_(z)) Since the motion of molecule is equally probable in all the three directions, therefore E_(K_(x)) = E_(K_(y)) = E_(K_(z)) =1/3 E_(K) =1/3 xx 3/2 kT = 1/2 kT , where k =R/N_(A) = Botzman constant. K.E. = 1/2 kT per molecule or =1/2 RT per mole. In vibration motion, molecules possess both kinetic energy as well as potential energy. This means energy of vibration involves two degrees of fiuedom. Vibration energy =2 xx 1/2kT =2 xx 1/2RT [ therefore two degrees of freedom per mole] If the gas molecules have n_(1) translational degrees of freedom, n_2 rotational degrees of freedom and n_(3) vibrational degrees of freedom, that total energy = n_(1)[(kT)/2] + n_(2) [(kT)/2] + n_(3) [(kT)/2] xx 2 Where 'n' is atomicity of gas. The rotational kinetic energy of H20 molecule is equal to

The value of the determinant Delta = |((1 - a_(1)^(3) b_(1)^(3))/(1 - a_(1) b_(1)),(1 - a_(1)^(3) b_(2)^(3))/(1 - a_(1) b_(2)),(1 - a_(1)^(3) b_(3)^(3))/(1 - a_(1) b_(3))),((1 - a_(2)^(3) b_(1)^(3))/(1 - a_(2) b_(1)),(1 - a_(2)^(3) b_(2)^(3))/(1 - a_(2) b_(2)),(1 - a_(2)^(3) b_(3)^(3))/(1 - a_(2) b_(3))),((1 - a_(3)^(3) b_(1)^(3))/(1 - a_(3) b_(1)),(1 - a_(3)^(3) b_(2)^(3))/(1 - a_(3) b_(2)),(1 - a_(3)^(3) b_(3)^(3))/(1 - a_(3) b_(3)))| , is

The following equilibria are given by : N_(2)+3H_(2) hArr 2NH_(3), K_(1) N_(2)+O_(2) hArr 2NO,K_(2) H_(2)+(1)/(2)O_(2) hArr H_(2)O, K_(3) The equilibrium constant of the reaction 2NH_(3)+(5)/(2)O_(2)hArr 2NO +3H_(2)O in terms of K_(1) , K_(2) and K_(3) is

Calculate the overall order of a reaction which has the rate expresison. (a) Rate = k[A]^((1)/(2))[B]^((3)/(2)) , (b) Rate = k[A]^((3)/(2))[B]^(-1)