Question

For the irreversible process, $A+B\to$products, the rate is first-order w.r.t. A and second-order w.r.t. B. If 1.0 mol each of A and B introduced into a 1.0 L vessel, and the initial rate was ${1.0\times 10}^{-2}$ mot $L^{-1}{s}^{-1}$, rate when half reactants have been turned into products is:

• A. ${1.25\times 10}^{-3}mol{L}^{-1}{s}^{-1}$
• B. ${1.0\times 10}^{-3}mol{L}^{-1}{s}^{-1}$
• C. ${2.50\times 10}^{-3}mol{L}^{-1}{s}^{-1}$
• D. ${2.0\times 10}^{-3}mol{L}^{-1}{s}^{-1}$

Answer 1

Answer: (A)

${1.25\times 10}^{-3}mol{L}^{-1}{s}^{-1}$

The Reaction is 1st order wrt A and 2nd order wrt B.

Initial number of moles of A and B are 1 mol and 1 mol respectively.

$Rate=k\left[A\right][B{]}^2$

When half the reactants are converted to products

No of moles of A =0.5 mol and B = 0.5 mol

The Rate when half reactions have been turned into products is,

$Rate=k\left[0.5\times A\right][0.5\times B{]}^2$

$Rate=\frac{1\times {10}^{-2}}{8}$ = $1.25\times {10}^{-3}mol{L}^{-1}{s}^{-1}$

Option A is correct