question_answer

  Identify, giving reasons, the faster reaction in each of the following: (i) \[2Na(s)+{{I}_{2}}(s)\xrightarrow{\,}\,2NaI(s)\] \[2Na(s)+{{I}_{2}}(g)\xrightarrow{\,}\,2NaI(s)\] (ii) \[C(s)+2{{H}_{2}}O(g)\xrightarrow{\,}\,2{{H}_{2}}(g)+C{{O}_{2}}(g)\] \[C(s)+2{{H}_{2}}O(s)\xrightarrow{Catalyst}\,2{{H}_{2}}(g)+C{{O}_{2}}(g)\] (iii) \[C(s)+1/2\,{{O}_{2}}(g)\xrightarrow{{{5}^{o}}0}\,CO(g)\] \[C(s)+1/2\,{{O}_{2}}(g)\xrightarrow{{{1}^{o}}0}\,CO(g)\] (iv)\[{{N}_{2}}\,(10\,\text{atm})+3{{H}_{2}}\,(10\,\,atm)\xrightarrow{\,}2N{{H}_{3}}(g)\] \[{{N}_{2}}(5\,atm)\,+3{{H}_{2}}\,(5\,atm)\,\xrightarrow{\,}\,2N{{H}_{3}}(g)\]

Answer:

                  (i) The second reaction is expected to be faster because iodine molecules in the gaseous state have greater surface area and more kinetic energy than the molecules in the solid state. (ii) The reaction carried in the presence of catalyst is faster because in the catalysed reaction the activation energy needed is less as compared to the uncatalysed reaction. (iii) The second reaction is faster because the increase in temperature increases the number of effective collisions among the reacting species. (iv) The first reaction will proceed at a faster speed because under high pressure the number of reactant molecules per unit volume are more compared with low pressure. This means that when the pressure is high, the collisions among the reactant molecules are more and as a result, the chemical reaction will be fast compared with low pressure.