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NEET 2010
Exam
NEET
Year
2010
Subject
Chemistry
Chapter
Chemical Kinetics

NEET 2010 · Chemistry · Chemical Kinetics

For the reaction ( N _ 2 O _ 5 ( g) arrow 2 NO _ 2 ( g)+(1 / 2) O _...

For the reaction (\mathrm{N}}_{2} \mathrm{O}}_{5}(g)→\mathrm{g}) \rightarrow2 \mathrm{NO}}_{2}( \mathrm{g})+(1 / 2) \mathrm{O}}_{2}(g)\mathrm{g})) the value of rate of disappearance of (\mathrm{N}}_{2} \mathrm{O}}_{5}) is given as (6.25 ×\times10^{-3}  mol L−1 s−1\mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}). The rate of formation of (\mathrm{NO}}_{2}) and (\mathrm{O}}_{2}) is given respectively as

Options

  1. A6.25×10−3 mol L−1 s−16.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 6.25×10−3 mol L−1 s−16.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
  2. B1.25×10−2 mol L−1 s−11.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 3.125×10−3 mol L−1 s−13.125 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
  3. C6.25×10−3 mol L−1 s−16.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 3.125×10−3 mol L−1 s−13.125 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
  4. D1.25×10−2 mol L−1 s−11.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 6.25×10−3 mol L−1 s−16.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

Correct answer

B. 1.25×10−2 mol L−1 s−11.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 3.125×10−3 mol L−1 s−13.125 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

Step-by-step solution

Key Idea Rate of disappearance of reactant = rate of appearance of product  or  −1 stoichiometric coefficient d [reactant ]dt of  reactant  =+1 stoichiometric d [product ]dt coefficient of  product  \begin{aligned}& \text{\text{ or } }-\frac{1}{\text{ stoichiometric coefficient }} \frac{\mathrm{d} \text{ [reactant }]}{\mathrm{dt}}\\ & \text{\text{ of }\text{ reactant } }\\ & =+\frac{1}{\text{ stoichiometric }} \frac{\mathrm{d} \text{ [product }]}{\mathrm{dt}}\\ & \text{ coefficient of \text{ product } }\\ & \end{aligned}For the reaction,N2O5( g)⟶2NO2( g)+12O2( g)−d[N2O5]dt=+12d[NO2]dt=+2 d[O2]dt∴d[NO2]dt=−2d[N2O5]dt=2×6.25×10−3 mol L−1 s−1=12.5×10−3 mol L−1 s−1=1.25×10−2 mol L−1 s−1\begin{aligned} & \mathrm{N}_2 \mathrm{O}_5(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \\ & \frac{-\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}=+\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}} \\ & =+\frac{2 \mathrm{~d}\left[\mathrm{O}_2\right]}{\mathrm{dt}} \\ & \therefore \quad \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}=-2 \frac{\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}} \\ & =2 \times 6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \\ & =12.5 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \\ & =1.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \\ & \end{aligned}d[O2]dt=−d[N2O5]dt×12=6.25×10−3 mol L−1 s−12=3.125×10−3 mol L−1 s−1\begin{aligned}\frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}& =-\frac{\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}} \times \frac{1}{2}\\ & =\frac{6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}}{2}\\ & =3.125 \times10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \end{aligned}
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