Question

For the reaction, $Cl_2 + 2I^- \longrightarrow \, I_2 + 2Cl^- ,$ the initial concentration of $I^-$ was 0.20 mol $L^{-1}$ and the concentration after 20 min was 0.18 mol $L^{-1}$. Then the rate of formation of $I_2$ in mol $L^{-1} \, min^{-1}$ would be

• A. $1 \times 10^{-3}$
• B. $5 \times 10^{-4}$
• C. $1 \times 10^{-4}$
• D. $2 \times 10^{-3}$
• E. $5 \times 10^{-3}$

Answer 1

Answer: (A)

Rate of formation of $I_2 = \frac{dI_2}{dt} = -\frac{1}{2} \frac{d [I^-]}{dt}$
$ = \frac{1}{2} \times \frac{0.20 - 0.18}{20} = \frac{1}{2} \times \frac{0.02}{20}$
$ = 5 \times 10^{-4} \, mol \, L^{-1} \, min^{-1}$