NEET UG 2010 — Chemistry Physical Chemistry

Key Idea Rate of disappearance of reactant $=$= rate of appearance of product $$\begin{aligned} & \text { or }-\frac{1}{\text { stoichiometric coefficient }} \frac{\mathrm{d} \text { [reactant }]}{\mathrm{dt}} \\ & \text { of reactant } \\ & =+\frac{1}{\text { stoichiometric }} \frac{\mathrm{d} \text { [product }]}{\mathrm{dt}} \\ & \text { coefficient of product } \\ & \end{aligned}$$​ or − stoichiometric coefficient 1​dtd [reactant ]​ of reactant =+ stoichiometric 1​dtd [product ]​ coefficient of product ​ For the reaction, $$\begin{aligned} & \mathrm{N}_2 \mathrm{O}_5(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \\ & \frac{-\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}=+\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}} \\ & =+\frac{2 \mathrm{~d}\left[\mathrm{O}_2\right]}{\mathrm{dt}} \\ & \therefore \quad \frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}=-2 \frac{\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}} \\ & =2 \times 6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \\ & =12.5 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \\ & =1.25 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \\ & \end{aligned}$$​N2​O5​( g)⟶2NO2​( g)+21​O2​( g)dt−d[N2​O5​]​=+21​dtd[NO2​]​=+dt2 d[O2​]​∴dtd[NO2​]​=−2dtd[N2​O5​]​=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​ $$\begin{aligned} \frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}} & =-\frac{\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}} \times \frac{1}{2} \\ & =\frac{6.25 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}}{2} \\ & =3.125 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \end{aligned}$$dtd[O2​]​​=−dtd[N2​O5​]​×21​=26.25×10−3 mol L−1 s−1​=3.125×10−3 mol L−1 s−1​