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Question: Nitrogen dioxide (\(N{{O}_{2}}\)) dissociates into nitric oxide (\(NO\)) and oxygen \(({{O}_{2}})\) ...

Nitrogen dioxide (NO2N{{O}_{2}}) dissociates into nitric oxide (NONO) and oxygen (O2)({{O}_{2}}) as follows:
2NO2−>2NO+O22N{{O}_{2}}->2NO+{{O}_{2}}
If the rate of decrease of concentration of NO2N{{O}_{2}} is 6.0×10−12mol L−1 s−16.0\times {{10}^{-12}}mol\,{{L}^{-1}}\,{{s}^{-1}}, what will be the rate of increase of concentration of O2{{O}_{2}}?
A.3×10−12mol L−1 s−13\times {{10}^{-12}}mol\,{{L}^{-1}}\,{{s}^{-1}}
B.6×10−12 mol L−1 s−16\times {{10}^{-12}}\,mol\,{{L}^{-1}}\,{{s}^{-1}}
C.1×10−12mol L−1 s−11\times {{10}^{-12}}mol\,{{L}^{-1}}\,{{s}^{-1}}
D.1.5×10−12mol L−1 s−11.5\times {{10}^{-12}}mol\,{{L}^{-1}}\,{{s}^{-1}}

Explanation

Solution

Try to write the reaction given in terms of the rate law equation, where a relation between the rate occurs for each of the reactants and products. Then, by using the given data in the question, we can find the required answer.

Complete step-by-step solution:
In order to answer our question, let us get to know the mathematical expression for writing the rates of reaction:
Let us consider the gaseous reaction between nitrogen dioxide and carbon monoxide:
NO2(g)+CO(g)−>CO2(g)+NO(g)N{{O}_{2}}(g)+CO(g)->C{{O}_{2}}(g)+NO(g)
In this case, as reactants and products appear in same stoichiometric proportions, therefore the rate of reaction may be expressed in terms of rate of disappearance of nitrogen dioxide or appearance of carbon dioxide , or nitric oxide, whichever may be convenient. So, the rate of reaction is mathematically given by:
  ⟹  −d[NO2]dt=−d[CO]dt=+d[CO2]dt=+d[NO]dt\implies-\dfrac{d[N{{O}_{2}}]}{dt}=-\dfrac{d[CO]}{dt}=+\dfrac{d[C{{O}_{2}}]}{dt}=+\dfrac{d[NO]}{dt}
However, in the reaction which is given in our question, 2NO2−>2NO+O22N{{O}_{2}}->2NO+{{O}_{2}},the stoichiometric coefficients are different. In such cases, we divide the rate of change of concentration by the stoichiometric coefficient of reactant or product that is involved in the reaction. Thus, we have:
  ⟹  −12d[NO2]dt=+12d[NO]dt=+d[O2]dt\implies-\dfrac{1}{2}\dfrac{d[N{{O}_{2}}]}{dt}=+\dfrac{1}{2}\dfrac{d[NO]}{dt}=+\dfrac{d[{{O}_{2}}]}{dt} (Eq i )
This is the required equation. Now, we have also been given that rate of decrease of concentration of NO2N{{O}_{2}} is 6.0×10−12mol L−1 s−16.0\times {{10}^{-12}}mol\,{{L}^{-1\,}}{{s}^{-1}}.

So, we have,
  ⟹  −d[NO2]dt=6.0×10−12mol L−1 s−1\implies-\dfrac{d[N{{O}_{2}}]}{dt}=6.0\times {{10}^{-12}}mol\,{{L}^{-1}}\,{{s}^{-1}}
Now, from equation ( i ) we have,
  ⟹  −12d[NO2]dt=d[O2]dt\implies-\dfrac{1}{2}\dfrac{d[N{{O}_{2}}]}{dt}=d\dfrac{[{{O}_{2}}]}{dt}
So,
  ⟹  d[O2]dt=12×d[NO2]dt   ⟹  =12×6.0×10−12=3.0×10−12 mol L−1 s−1 \begin{aligned} &\implies \dfrac{d[{{O}_{2}}]}{dt}=\dfrac{1}{2}\times \dfrac{d[N{{O}_{2}}]}{dt} \\\ &\implies =\dfrac{1}{2}\times 6.0\times {{10}^{-12}}=3.0\times {{10}^{-12\,}}mol\,{{L}^{-1}}\,{{s}^{-1}} \\\ \end{aligned}

So, we obtain the rate of appearance of oxygen as 3.0×10−12 mol L−1 s−13.0\times {{10}^{-12}}\,mol\,{{L}^{-1}}\,{{s}^{-1}}. Hence, the correct will be (A).

Note: It is to be noted that as the question has asked for the rate of appearance, there is a positive sign. Also, the rate of appearance of NO2N{{O}_{2}} would have been negative, but we are talking about the rate of disappearance, so the sign is reversed.