For the second-order reaction: A + B - Products, the rate constant, k, is given as k = 2.303/((a-b)t) log(b(a-x) / a(b-x)), where a and b are the initial concentrations of 'A' and 'B' and x is the ch

Question

For the second-order reaction: A + B -> Products, the rate constant, k, is given as k = 2.303/((a-b)t) log(b(a-x) / a(b-x)), where a and b are the initial concentrations of 'A' and 'B' and x is the change in concentration after time t. If b >> a, the reaction reduces to

Prashant Kotian · Accepted Answer

Correct answer: A. If b >> a, the concentration of B remains practically constant (b-x ≈ b and b-a ≈ b). Substituting these into the second-order rate equation: k ≈ 2.303/(b*t) * log((a-x)/a) => k*b = 2.303/t * log(a/(a-x)). Let k' = k*b (pseudo-first-order rate constant). This represents a first-order rate law with respect to A.

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