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Question: \(N{{O}_{2}}\) required for a reaction is produced by the decomposition of \({{N}_{2}}{{O}_{5}}\) in...

NO2N{{O}_{2}} required for a reaction is produced by the decomposition of N2O5{{N}_{2}}{{O}_{5}} in CCl4CC{{l}_{4}} as per the equation:
2N2O5(g)→4NO2(g)+O2(g)2{{N}_{2}}{{O}_{5}}(g)\to 4N{{O}_{2}}(g)+{{O}_{2}}(g)
The initial concentration of N2O5{{N}_{2}}{{O}_{5}} is 3.00 mol L−13.00\,mol\,{{L}^{-1}} and it is 2.75 mol L−12.75\,mol\,{{L}^{-1}} after 30 minute. The rate of formation of NO2N{{O}_{2}} is:
A.2.083×10−3 mol L−1 min⁡−12.083\times {{10}^{-3}}\,mol\,{{L}^{-1\,}}{{\min }^{-1}}
B.4.167×10−3 mol L−1 min⁡−14.167\times {{10}^{-3}}\,mol\,{{L}^{-1}}\,{{\min }^{-1}}
C.8.333×10−3 mol L−1min⁡−18.333\times {{10}^{-3\,}}mol\,{{L}^{-1}}{{\min }^{-1}}
D.1.667×10−2 mol L−1min⁡−11.667\times {{10}^{-2}}\,mol\,{{L}^{-1}}{{\min }^{-1}}

Explanation

Solution

. Recall the rate law equation. Try to express the chemical equation given in the question in terms of the rate for individual reactants and products. Also, use the concentrations and time given above to find the rate and use the rate to find the required answer.

Complete step by step answer:
In chemical kinetics, the rate of a chemical reaction is the ratio of the difference in concentration and the time taken to reach the final concentration. The rate of a reaction can also be called the speed of a reaction, which gives us an idea about how fast the reaction occurs. Now, if the rate of the reaction is positive, that means, concentration is decreased in due course of time and generally it happens with the reactants, as they get converted into products. A negative rate implies that concentration is increasing over time. Now, let us write the equation for the rate of appearance and disappearance of the reactants and products for the chemical reaction:
−12d[N2O5]dt=+14d[NO2]dt=+d[O2]dt-\dfrac{1}{2}\dfrac{d[{{N}_{2}}{{O}_{5}}]}{dt}=+\dfrac{1}{4}\dfrac{d[N{{O}_{2}}]}{dt}=+\dfrac{d[{{O}_{2}}]}{dt}…………(1)

Here, as the reactant is getting used up, so a negative sign is introduced, and vice versa for the products. Also, the stoichiometric coefficients are reversed, as it is the rule.
Now in order to find the disappearance of N2O5{{N}_{2}}{{O}_{5}}, we can write
d[N2O5]dt=3−2.7530 =56×10−2M min⁡−1 \begin{aligned} & \dfrac{d[{{N}_{2}}{{O}_{5}}]}{dt}=\frac{3-2.75}{30} \\\ & =\dfrac{5}{6}\times {{10}^{-2}}M\,{{\min }^{-1}} \\\ \end{aligned}

Now, using this value in equation (1) we have,
d[NO2]dt=2×56×10−2 =1.667×10−2 \begin{aligned} & \dfrac{d[N{{O}_{2}}]}{dt}=2\times \dfrac{5}{6}\times {{10}^{-2}} \\\ & =1.667\times {{10}^{-2}} \\\ \end{aligned}
So, the rate of formation of NO2N{{O}_{2}} is 1.667×10−2 mol L−1 min⁡−11.667\times {{10}^{-2}}\,mol\,{{L}^{-1}}\,{{\min }^{-1}},
So, the correct answer is “Option D”.

Note: As the question had asked us to find the rate of formation, we did not put the negative sign in our answer. In case the question had asked us about the rate of disappearance, our answer would have been −1.667×10−2 mol L−1 min⁡−1-1.667\times {{10}^{-2}}\,mol\,{{L}^{-1}}\,{{\min }^{-1}}.