Question

The reaction between ${{Cr}_2{O}_7}^{2-}$ and $HN{O}_2$ in an acidic medium is:
${{Cr}_2{O}_7}^{2-}+5H^{+}+3HN{O}_2\to 2{Cr}^{3+}+3N{O_3}^{-}+4H_{2}O$
The rate of disappearance of ${Cr}_2{O_7}^{2-}$ is found to be $2.4\times {10}^{-4}$ $mol$ $L^{-1}$ $s^{-1}$ during a measured time interval. Find the rate of disappearance of $HN{O}_2$ and the rate of appearance of ${Cr}^{3+}$ during this time interval.

Answer 1

The equality in this case is:
$-\frac{\Delta \left[{Cr}_2{O}_7^{2-}\right]}{\Delta t}=-\frac{1}{3}\frac{\Delta \left[HN{O}_2\right]}{\Delta t}=\frac{1}{2}\frac{\Delta \left[{Cr}^{3+}\right]}{\Delta t}$
It is given that,
$-\frac{\Delta \left[{Cr}_2{O}_7^{2-}\right]}{\Delta t}=2.4\times {10}^{-4}$ $mol$ $L^{-1}$ $s^{-1}$
So, $-\frac{\Delta \left[HN{O}_2\right]}{\Delta t}=3\times 2.4\times {10}^{-4}=7.2\times {10}^{-4}$ $mol$ $L^{-1}$ $s^{-1}$
and $\frac{\Delta \left[{Cr}^{3+}\right]}{\Delta t}=2\times 2.4\times {10}^{-4}=4.8\times {10}^{-4}$ $mol$ $L^{-1}$ $s^{-1}$