The equilibrium concentration of `[B]_(e)` for the reversible reaction `A harrB` can be evaluated by the expression:
`K_(c)[A]_(e)^(-1)`
`(k_(f))/(k_(b))[A]_(e)^(-1)`
`k_(f)k_(b)^(-1)[A]_(e)`
`k_(f)k_(b)[A]_(e)^(-1)`

The equilibrium constant K_(p) for the following rection at 191^(@)C is 1.24. what is K_(c) ? B(s)+(3)/(2)F_(2)(g)hArrBF_3(g)

For and elementary reaction 2A underset(k_(2))overset(k_(1))hArr B , the rate of disappearance of A iss equal to (a) (2k_(1))/(k_(2))[A]^(2) (b) -2k_(1)[A]^(2) + 2k_(2)[B] ( c) 2k_(1)[A]^(2) - 2k_(2)[B] (d) (2k_(1) - k_(2))[A]

For a reaction of reversible nature, net rate is ((dx)/(dt))=k_(1)[A][B]-k_(2)[C] hence, given reaction is :

For reaction A rarr B , the rate constant K_(1)=A_(1)(e^(-E_(a_(1))//RT)) and the reaction X rarr Y , the rate constant K_(2)=A_(2)(e^(-E_(a_(2))//RT)) . If A_(1)=10^(9) , A_(2)=10^(10) and E_(a_(1)) =1200 cal/mol and E_(a_(2)) =1800 cal/mol , then the temperature at which K_(1)=K_(2) is : (Given , R=2 cal/K-mol)

The value of int_0^1lim_(n->oo)sum_(k =0)^n(x^(k+2)2^k)/(k!)dx is: (a) e ^(2) -1 (b) 2 (c) (e ^(2)-1)/(2) (d) (e ^(2) -1)/(4)

For a reaction of reversible nature , net rate is ((dx)/(dt))=k_(1)[A]^(2)[B]^(1)-k_(2)[C] , hence , given reactionis:

For a reaction A(s)+2B^(o+) rarr A^(2+)+2B K_(c) has been found to be 10^(12) . The E^(c-)._(cell) is

Seven rods A, B, C, D, E, F and G are joined as shown in figure. All the rods have equal cross-sectional area A and length l. The thermal conductivities of the rods are K_(A)=K_(C)=K_(0) , K_(B)=K_(D)=2K_(0) , K_(E)=3K_(0) , K_(F)=4K_(0) and K_(G)=5K_(0) . The rod E is kept at a constant temperature T_(1) and the rod G is kept at a constant temperature T_(2) (T_(2)gtT_(1)) . (a) Show that the rod F has a uniform temperature T=(T_(1)+2T_(2))//3 . (b) Find the rate of heat flowing from the source which maintains the temperature T_(2) .

Which of the following expression for % dissociation of a monoacidic base (BOH) in aqueous solution at appreciable concentration is not correct? (a) 100xxsqrt((K_(b))/(c)) (b) (1)/(1+10^((pK_(b)-pOH)) (c) (K_(w)[H^(+)])/(K_(b)+K_(w)) (d) (K_(b))/(K_(b)+[OH^(-)])

If inte^(x)/(x^(n))dx and -e^(x)/(k_(1)x^(n-1))+1/(k_(2)-1)I_(n-1) , then (k_(2)-k_(1)) is equal to: