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

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$1.0 \times 10^{- 2} {mol L}^{- 1} s^{-}$ ¹

$2.50 \times 10^{- 3} {mol L}^{- 1} s^{-}$ ¹

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

$2.0 \times 10^{- 2} {mol L}^{- 1} s^{-}$ ¹

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} A + B \to product \\ r = K {[A]}^{1} {[B]}^{2} \\ r_{1} = K {[1]}^{1} {[1]}^{2} = 1 \times 10^{- 2} (K = 1 \times 10^{- 2}) \\ r_{1} = K [\frac{1}{2}] {[\frac{1}{2}]}^{2} = 1 \times 10^{- 2} \times \frac{1}{8} \\ = 1.25 \times 10^{- 3} {mol L}^{- 1} S^{- 1} \end{array}$