Question

Which of the following change will shift the reaction towards the product?
$I_2\left(g\right)\rightleftharpoons 2I\left(g\right),\Delta H_r^{\circ }\left(298K\right)=+150kJ$

• A. Increase in concentration of $I\left(g\right)$
• B. Decrease in concentration of $I_2\left(g\right)$
• C. Increase in temperature
• D. Increase in total pressure

Answer 1

The correct option is C Increase in temperature

The reaction is $I_2\left(g\right)\rightleftharpoons 2I\left(g\right),\Delta H_r^{\circ }\left(298K\right)=+150kJ$.

According to Le-Chatelier's principle, the conditions favouring the formation of $I$ are given below:
(i) High concentration of $I_2\left(g\right)$.
(ii) High temperature (reaction is endothermic so, on increasing the temperature equilibrium it will shift in the direction where heat is absorbed).
(iii) Low pressure (moles of products are more than moles of reactant. Therefore, if pressure is increased, equilibrium will shift in the direction where moles are decreasing).