Question

Of the following which change will shift the reaction towards the product? $I_2(g) \rightleftharpoons 2I(g)$ $\Delta H^{\circ} (298\, K) = +150\, kJ$

• A. Increase in concentration of $I$
• B. Decrease in concentration of $I_2$
• C. Increase in temperature
• D. Increase in total pressure

Answer 1

Answer: (C)

Increase in temperature

Answer 2

$I _{2}(g) 2 I (g) ; \Delta H_{r}^{\circ}(298\, K )=+150\, kJ$ According to Le-Chatelier's principle the conditions favouring formation of $I$ are (i) High concentration of $I _{2}$ (ii) High temperature (reaction is endothermic so, on increasing the temperature equilibrium will shift in the direction where heat is absorbed). (iii) Low pressure (moles of products are more than moles of reactant, so, if pressure is increased equilibrium will shift in the direction where volume is less).