NEET UG 2009 — Chemistry Physical Chemistry

Key Idea Rate of disappearance of reactant $=-\frac{1}{\text { stoichiometry }} \times \frac{\text { change in conc of reactant }}{\text { time taken }}$=− stoichiometry 1​× time taken  change in conc of reactant ​ coefficient of the reactant Rate of appearance of product change in conc $=\frac{1}{\text { stoichiometry coefficient }} \times \frac{\text { of product }}{\text { time taken }}$= stoichiometry coefficient 1​× time taken  of product ​ For the reaction, $$\begin{aligned} & \mathrm{N}_2+3 \mathrm{H}_2 \longrightarrow 2 \mathrm{NH}_3 \\ & \begin{aligned} & \text { Rate }=-\frac{\mathrm{d}\left[\mathrm{N}_2\right]}{\mathrm{dt}}=-\frac{1}{3} \frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=+\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{NH}_3\right]}{\mathrm{dt}} \\ & \text { or } \quad-\frac{1}{3} \frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=+\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{NH}_3\right]}{\mathrm{dt}} \\ &-\frac{\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=\frac{3}{2} \times 2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \\ &=3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \end{aligned} \end{aligned}$$​N2​+3H2​⟶2NH3​​ Rate =−dtd[N2​]​=−31​dtd[H2​]​=+21​dtd[NH3​]​ or −31​dtd[H2​]​=+21​dtd[NH3​]​−dtd[H2​]​=23​×2×10−4 mol L−1 s−1=3×10−4 mol L−1 s−1​​