Question

For a process, $A+B\to$ products, the rate of the reaction is second order with respect to A and zero order with respect to B. when 1.0 mole each of A and B are taken in a one litre vessel, the initial rate is $1\times {10}^{-2}$ mol $L^{-1}{s}^{-1}$ . The rate of the reaction, when 50% of the reactants have been converted to products would be:

• A. $1\times {10}^{-2}$ mol $L^{-1}{s}^{-1}$
• B. $2.5\times {10}^{-3}$ mol $L^{-1}{s}^{-1}$
• C. $5.0\times {10}^{-2}$ mol $L^{-1}{s}^{-1}$
• D. $0.5\times {10}^{-2}$ mol $L^{-1}{s}^{-1}$

Answer 1

The correct option is B $2.5\times {10}^{-3}$ mol $L^{-1}{s}^{-1}$

The rate law expression is $r=k\left[A{]}^2\right[B{]}^0.$
The initial rate is $1\times {10}^{-2}.$
Hence, $1\times {10}^{-2}=k\left[1{]}^2\right[1{]}^0.$
$k=1\times {10}^{-2}$
when 50% reactants are converted into products then amount of reactant will become 0.5.
Thus, $r=k\left[A{]}^2\right[B{]}^0=1\times {10}^{-2}\left[0.5{]}^2\right[0.5{]}^0=2.5\times {10}^{-3}.$