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In a certain reactionB^(n+)is getting converted toB^((n+4)+) in solution. The rate constant for the reaction is measured by titrating a volume of the solution with a reducing agent which reacts only with B^(n+)andB^((n+4)+) . In this process it converts.B^(n+)" to "B^((n-2)+)andB^((n+4)+)" to "B^((n-1)+). Att = 0 ,the volume of reagent consumed is 25mL and at t = 10min , the volume used is 32mL. Calculate the rateconstant for conversion ofB^((n)+)" to "B^((n+4)+)assuming it to be first order reaction. |
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Answer» Solution :According to the given data `B^((N)+)+2erarrB^((n-2)+)""...[1]` `B((n+n)+)+5erarrB^((n-1)+)""...[2]` `""B^(n+)RARRB^((n+4)+)` `{:("Initial number of moles:",a,0),("Number of moles at time":,(a-x),x):}` From reaction [1] , 1 MOL of `B^(n+)-=2xx1 " EQUIVALENT of "B^(n+)` `a" mol of "B^(n+)-=2xxa " equivalent of "B^(n+)` and ( a-x) mol of `B^(n+)-=2xx(a -x)" equivalent of "B^(n+)` Let the normality of the reducing AGENT be N. Hence , for reaction [1], `axx2=Nxx25[thereforeN_(1)V_(1)=N_(2)V_(2)]` or,`2a=25N thereforea (25)/(2)(N)` From reaction [2] : `1"mol of "B^(n+4)-=5xx1"equiv. of "B^((n+4)+)` `thereforex"mol of "B^(n+4)-=5x" equivalent of"B^((n+4)+)` Hence , for reaction [2] ,`(a-x)xx2+5x=32xx(N)` or,`2a+3x=32(N)` or,25(N) +3x=32N[ putting 2a = 25 (N)] or, 3x = 7 (N)or,`x=(7)/(3)(N)` For a first order reaction , rate constant `k=(2.303)/(t)log.([A]_(0))/([A])` or, `k =(2.303)/(10)log.(a)/(a-x) " or, " k = (2.303)/(10)log.((25)/(2)(N))/(((25)/(2)-(7)/(3))(N))` `thereforek = 2.07xx10^(-2)"min"^(-1)` `therefore`The rate constant of the reaction`= 2.07xx10^(-2)"min"^(-1)` |
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