for the reaction a+b > products , the rate is first order wrt a and second order wrt to b . if 1 mol each of a and b were introduced into a 1 l vessel and the initial rate were 1x10^-2. calculate the rate when half the reactants have been turned into product

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According to the question, 
the reaction can be written as:

A+B productthe reaction is first order wrt A and second order wrt BInitial number of moles of A = 1 mol and B= 1 molInitial volume = 1LInitial rate is 10-2Thus the rate equation can be written as:Rate = k [a]1[b]2where k is the rate constantSince the initial concentration is 1mol/L for both Aand Bthus k = 10-2  mol-3L3Now when halk the reactants have been converted to the products,a = 0.5 mol, b = 0.5molthus the rate will be:Rate = k[0.5mol/l][0.5molL]2Rate = k/8= 10-28Regards

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the final ans is 1.25 x 10-3
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The moles left of both the compond will be .5moles and because the B is of secound order it will be .00125
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