Nitrogen dioxide gas dissociates to give nitric oxide and oxygen. For this reaction, when a graph is plotted between concentration of ${NO}_{2}$ and time, the slope of the tangent drawn at time $t = 10 s$ is found to be $6.8 \times 10^{- 4} mol L^{- 1} s^{- 1}$ . Predict the slope of the graph at $t = 10 s$ when the concentration of $NO$ and $O_{2}$ are plotted against time. Justify your answer.

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Solution

Nitrogen dioxide gas dissociates to give nitric oxide and oxygen. For this reaction, when a graph is plotted between the concentration of ${NO}_{2}$ and time, the slope of the tangent drawn at time $t = 10 s$ is found to be $6.8 \times 10^{- 4} mol L^{- 1} s^{- 1}$
The following is how the reaction goes:
$2 {NO}_{2} (g) \to 2 NO (g) + O_{2} (g) Nitrogen Nitrogen Oxygen dioxide Oxide$
Now consider the tangent's slope. $\frac{d [{NO}_{2}]}{dt}$ $= 6.8 \times$ $10^{- 4} mol L^{- 1} s^{- 1}$ at $t = 10 s$ .
Now, rate of the given reaction can be expressed as- $rate = - \frac{1}{2} \frac{d [{NO}_{2}]}{dt} = \frac{1}{2} \frac{d [NO]}{dt} = \frac{d [O_{2}]}{dt}$
The slope of the graph at $t = 10 s$ when the concentration $NO$ is plotted against time: $Slope = \frac{d [NO]}{dt} = - \frac{d [{NO}_{2}]}{dt} = - 6.8 \times 10^{- 4} mol . L^{- 1} . s^{- 1}$
The slope of the graph at $t = 10 s$ when the concentration $O_{2}$ is plotted against time: $Slope = \frac{d [O_{2}]}{dt} = - \frac{1}{2} \frac{d [{NO}_{2}]}{dt} = \frac{- 6.8 \times 10^{- 4} mol . L^{- 1} . s^{- 1}}{2} Slope = - 3.4 \times 10^{- 4} mol . L^{- 1} . s^{- 1}$

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