Question

From the given table for reaction between A and B:
$\begin{array}{ccccc}& & & \text{Initial rate}& \text{Initial rate}\\ & \left[A\right]{\text{mol litre}}^{-1}& \left[B\right]{\text{mol litre}}^{-1}& \text{at 300 K}& \text{at 320 K}\\ & & & {\text{mol litre}}^{-1}{\text{s}}^{-1}& {\text{mol litre}}^{-1}{\text{s}}^{-1}\\ \text{(I)}& 2.5\times {10}^{-4}& 3.0\times {10}^{-5}& 5.0\times {10}^{-4}& 2.0\times {10}^{-3}\\ \text{(II)}& 5.0\times {10}^{-4}& 6.0\times {10}^{-5}& 4.0\times {10}^{-3}& -\\ \text{(III)}& 1.0\times {10}^{-3}& 6.0\times {10}^{-5}& 1.6\times {10}^{-2}& -\end{array}$

Rate constant at $300\text{K}$ is $x\times {10}^8{\text{mol}}^{-2}{\text{litre}}^2{\text{s}}^{-1}$. Find the value of $15x$.

• A. 4
• B. 2.66
• C. 4.0
• D. 4.00
• E. 2.67

Answer 1

Answer: (A), (B), (C), (D), (E)

Let the order with respect to A is $x$ and order with respect to B is $y$.
Then, $\text{rate}=k\left[A{]}^x\right[B{]}^y$
$5.0\times {10}^{-4}=k[2.5\times {10}^{-4}{]}^x[3.0\times {10}^{-5}{]}^y..\left(i\right)4.0\times {10}^{-3}=k[5.0\times {10}^{-4}{]}^x[6.0\times {10}^{-5}{]}^y..\left(ii\right)1.6\times {10}^{-2}=k[1.0\times {10}^{-3}{]}^x[6.0\times {10}^{-5}{]}^y..\left(iii\right)$
Deviding equation (ii) by (iii), we get
$0.25=[5.0\times {10}^{-1}{]}^x$
$\Rightarrow x=2$
Deviding equation $\left(i\right)$ by $\left(ii\right)$ we get,
$0.125=\left(0.5{)}^x\right(0.5{)}^y$
$\Rightarrow 0.125=(0.5{)}^{x+Y}$
$\therefore x+y=3$
$\Rightarrow Y=1$

$\therefore$ The rate equation for the reaction is $\text{rate}=k\left[A{]}^2\right[B]$

Rate constant $\left(k_1\right)$ at $300\text{K}$,
$k_1=\frac{\text{Rate}}{\left[A{]}^2\right[B]}=\frac{5.0\times {10}^{-4}{\text{mol litre}}^{-1}{\text{s}}^{-1}}{(2.5\times {10}^{-4}{\text{mol litre}}^{-1}{)}^2(3.0\times {10}^{-5}{\text{mol litre}}^{-1})}=\frac{2}{2.5\times 3}\times {10}^8{\text{mol}}^{-2}{\text{litre}}^2{\text{s}}^{-1}$
$\therefore x=\frac{2}{2.5\times 3}$
Hence, $15x=4$