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 $1.0mol$ each of $A$ and $B$ is introduced into a $1.0L$ 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:

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

Answer 1

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

Now rate of reaction (r) is given by

$r=k\left[A{]}^1\right[B{]}^2$

When $\left[A\right]=1m$ $\left[B\right]=1m$

then $r={10}^{-2}M{s}^{-1}$

$k={10}^{-2}{M}^{-2}{s}^{-1}$

Now,

$\left[A\right]=\frac{1}{2}M$ $\left[B\right]=\frac{1}{2}M$.....(1)

$\therefore r={10}^{-2}\left(\frac{1}{2}\right).{\left(\frac{1}{2}\right)}^2$

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

Equation (1) comes from first that half the reaction is completed.