Question

For the non-equilibrium 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 $1.0\times 1{\text{0}}^{\text{- 2}}\text{mol}{\text{L}}^{\text{- 1}}{\text{s}}^{\text{- 1}}$. Find the rate when half reactants have been turned into products.

• A. $1.25\times {10}^{-3}\text{mol}{\text{L}}^{\text{- 1}}{\text{s}}^{\text{- 1}}$
• B. $1.0\times {10}^{-2}\text{mol}{\text{L}}^{\text{- 1}}{\text{s}}^{\text{- 1}}$
• C. $2.50\times {10}^{-3}\text{mol}{\text{L}}^{\text{- 1}}{\text{s}}^{\text{- 1}}$
• D. $2.0\times {10}^{-2}\text{mol}{\text{L}}^{\text{- 1}}{\text{s}}^{\text{- 1}}$

Answer 1

The correct option is A $1.25\times {10}^{-3}\text{mol}{\text{L}}^{\text{- 1}}{\text{s}}^{\text{- 1}}$

$V_o=k\left[A_o\right][B_o{]}^2$ $\Rightarrow k=\frac{V_o}{\left[A_o\right][B_o{]}^2}$ $=\frac{1\times {10}^{-2}}{1}=1\times {10}^{-2}$

When both $\left[A_f\right]=\left[A_o/2\right]$ and $\left[B_f\right]=\left[B_o/2\right]$

$V_f=k\left[A_f\right][B_f{]}^2=1\times {10}^{-2}\times 0.125=1.25\times {10}^{-3}mol{L}^{-1}{s}^{-1}$