1.

A reaction is of first order in reactant A and of second order in reactant B. How is the rate of this reaction affected when (i) the concentration of B alone is increased to three times. (ii) the concentrations of A as well as B are doubled?

Answer»

Solution :`(dx)/(DT)= k[A][B]^(2) " Let "[A]=a, [B]=b, (dx)/(dt)=k.a.b^(2)`
(i) When the concentration of B is increased to THREE TIMES,
`(dx)/(dt)= ka(3b)^(2) or (dx)/(dt)=9kab^(2)`
Thus the rate increase 9 times.
(ii) When the concentration of A as well as B are DOUBLED,
`(dx)/(dt)= k(2a)(2b)^(2)" or "(dx)/(dt)=k.2a.4b^(2) " or "(dx)/(dt)=8kab^(2)`
Thus the rate increases 8 times.


Discussion

No Comment Found

Related InterviewSolutions