Kinetics of chemical reactions
Question

In the reversible reaction $2\text{\hspace{0.17em}}N{O}_{2}⇌{N}_{2}{O}_{4}$ The rate of dis appearance of $N{O}_{2}$  is equal to

Moderate
Solution

For the given reversible reaction, $2\text{\hspace{0.17em}}N{O}_{2}\underset{{k}_{2}}{\overset{{k}_{1}}{⇌}}{N}_{2}{O}_{4}$ The rate of disapperane of ${\text{NO}}_{2}$ =?Let the reaction be,$2\text{\hspace{0.17em}}N{O}_{2}\underset{{k}_{2}}{\overset{{k}_{1}}{⇌}}{N}_{2}{O}_{4}$ For the forward reaction rate $\left({r}_{f}\right)=\text{\hspace{0.17em}}-\frac{1}{2}\frac{d\left[N{O}_{2}\right]}{dt}=\frac{d\left[{N}_{2}{O}_{4}\right]}{dt}={K}_{1}{\left[N{O}_{2}\right]}^{2}$ For backward reaction Rate $=-\frac{d\left[{N}_{2}{O}_{4}\right]}{dt}=\frac{1}{2}\frac{d\left[N{O}_{2}\right]}{dt}={K}_{2}\left[{N}_{2}{O}_{4}\right].$ Rate of disappearance of $N{O}_{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ (Rate of disappearance of $N{O}_{2}-$  Rate of appearance of $N{O}_{2}$ )$=2{K}_{1}{\left[N{O}_{2}\right]}^{2}-2{K}_{2}\left[{N}_{2}{O}_{4}\right].$

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