For the reversible reaction A `harr` B at equilibrium `([B])/([A])` = `6times10^(4)`
-`(d[A])/(dt)` = `3times10^(6)s^(-1)`[A]
-`(d[B])/(dt)`=K[B] . Thus K is

For the reaction: aA + bB rarr cC+dD Rate = (dx)/(dt) = (-1)/(a)(d[A])/(dt) = (-1)/(b)(d[B])/(dt) = (1)/( c)(d[C])/(dt) = (1)/(d)(d[D])/(dt) In the following reaction, xA rarr yB log.[-(d[A])/(dt)] = log.[(d[B])/(dt)] + 0.3 where negative isgn indicates rate of disappearance of the reactant. Thus, x:y is:

If for the reaction A to B" rate" =-(d[A])/(dt)=2(d[B])/(dt) then, rate law is

For the reaction: aA + bB rarr cC+dD Rate = (dx)/(dt) = (-1)/(a)(d[A])/(dt) = (-1)/(b)(d[B])/(dt) = (1)/( c)(d[C])/(dt) = (1)/(d)(d[D])/(dt) A reaction follows the given concentration-time graph. The rate for this reaction at 20 s will be

For the reaction: aA + bB rarr cC+dD Rate = (dx)/(dt) = (-1)/(a)(d[A])/(dt) = (-1)/(b)(d[B])/(dt) = (1)/( c)(d[C])/(dt) = (1)/(d)(d[D])/(dt) For reaction 3BrO^(ɵ) rarr BrO_(3)^(ɵ) + 2Br^(ɵ) , the value of rate constant at 80^(@)C in the rate law for -(d[BrO^(ɵ)])/(dt) was found to be 0.054 L mol^(-1)s^(-1) . The rate constant (k) for the reaction in terms of (d[BrO_(3)^(ɵ)])/(dt) is

A+ 2B rarr C+D." If "-(d[A])/(dt)=5xx10^(-4)" mol L"^(-1) s^(-1)," then "-(d[B])/(dt) is :

In the reaction , A+2Bto6C+2D , if the initial rate -(d[A])/(dt) at t= 0 is 2.6xx10^(-2)Msec^(-1) , what will be the value of -(d[B])/(dt) at t=0?