SolveIt is under development
SolveItJEE Main
Chemistry · 2019

JEE Main · 10 April 2019, Shift 2 · Q59

For the reaction of H_2 with I_2, the rate constant is 2.5 × 10^-4 dm^3 mol^-1 s^-1 at 327 °C and 1.0 dm^3 mol^-1 s^-1 at 527^° C. The activation…

For the reaction of $\displaystyle \mathrm{H}_2$ with $\displaystyle \mathrm{I}_2$, the rate constant is $\displaystyle 2.5 \times 10^{-4} \mathrm{dm}^3 \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$ at $\displaystyle 327$ °C and $\displaystyle 1.0 \mathrm{dm}^3 \mathrm{~mol}^{-1} \mathrm{~s}^{-1}$ at $\displaystyle 527^{\circ} \mathrm{C}$. The activation energy for the reaction, in $\displaystyle \mathrm{kJ} \mathrm{mol}^{-1}$ is : $$\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) $$
ShareWhatsAppTelegram

More from Chemical Kinetics

JEE Main 2019 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.