Question

The reaction between $C{r}_2{O}_7^{2-}$ and $HN{O}_2$ in an acidic medium is :
$C{r}_2{O}_7^{2-}+5H^{+}+3HN{O}_2\to 2C{r}^{3+}+3N{O}_3^{-}+4H_{2}O$
The rate of disappearance of $C{r}_2{O}_7^{2-}$ is found to be $2.4\times {10}^{-4}\text{mol}{L}^{-1}{s}^{-1}$ during a measured time interval. Find the rate of disappearance of $HN{O}_2$ and the rate of appearance of $C{r}^{3+}$ during this time interval respectively.

• A. $7.2\times {10}^{-4}\text{mol}{L}^{-1}{s}^{-1}$ and $4.8\times {10}^{-4}\text{mol}{L}^{-1}{s}^{-1}$
• B. $4.8\times {10}^{-4}\text{mol}{L}^{-1}{s}^{-1}$ and $7.2\times {10}^{-4}\text{mol}{L}^{-1}{s}^{-1}$
• C. $2.4\times {10}^{-4}\text{mol}{L}^{-1}{s}^{-1}$ and $2.4\times {10}^{-4}\text{mol}{L}^{-1}{s}^{-1}$
• D. $4.8\times {10}^{-4}\text{mol}{L}^{-1}{s}^{-1}$ and $2.4\times {10}^{-4}\text{mol}{L}^{-1}{s}^{-1}$

Answer 1

The correct option is A $7.2\times {10}^{-4}\text{mol}{L}^{-1}{s}^{-1}$ and $4.8\times {10}^{-4}\text{mol}{L}^{-1}{s}^{-1}$

For the given reaction $C{r}_2{O}_7^{2-}+5H^{+}+3HN{O}_2\to 2C{r}^{3+}+3N{O}_3^{-}+4H_{2}O$
The equality is :
$-\frac{\Delta \left[C{r}_2{O}_7^{2-}\right]}{\Delta t}=-\frac{1}{3}\frac{\Delta \left[HN{O}_2\right]}{\Delta t}=\frac{1}{2}\frac{\Delta \left[C{r}^{3+}\right]}{\Delta t}$
It is given that,
$-\frac{\Delta \left[C{r}_2{O}_7^{2-}\right]}{\Delta t}=2.4\times {10}^{-4}\text{mol}{L}^{-1}{s}^{-1}$
So, $-\frac{1}{3}\frac{\Delta \left[HN{O}_2\right]}{\Delta t}=2.4\times {10}^{-4}\text{mol}{L}^{-1}{s}^{-1}$
$-\frac{\Delta \left[HN{O}_2\right]}{\Delta t}=3\times 2.4\times {10}^{-4}=7.2\times {10}^{-4}\text{mol}{L}^{-1}{s}^{-1}$
and $\frac{1}{2}\frac{\Delta \left[C{r}^{3+}\right]}{\Delta t}=2.4\times {10}^{-4}$
$\frac{\Delta \left[C{r}^{3+}\right]}{\Delta t}=2\times 2.4\times {10}^{-4}=4.8\times {10}^{-4}\text{mol}{L}^{-1}{s}^{-1}$