A mixture of 0.75 mol of N_(2) and 1.20 mol of H_(2) is placed in a 3 L container. When the reaction N_(2)+3H_(2) hArr2NH_(3) reaches the equilibrium, the concentration of H_(2) is 0.1 M. Calculate the concentration of N_(2) and NH_(3) when the reaction is carried out with double the number of moles.

A mixture of 0.75 mol of N_(2) and 1.20 mol of H_(2) is placed in a 3

1.	A mixture of 0.75 mol of N_(2) and 1.20 mol of H_(2) is placed in a 3 L container. When the reaction N_(2)+3H_(2) hArr2NH_(3) reaches the equilibrium, the concentration of H_(2) is 0.1 M. Calculate the concentration of N_(2) and NH_(3) when the reaction is carried out with double the number of moles.
Answer» Solution :`N_(2) + 3H_(2) hArr 2NH_(3)` Concentration of `N_(2)=0.75//3=0.25` mol `L^(-1)` Concentration of `H_(2) =1.20//3=0.4 ` mol `L^(-1)` `N_(2) + 3H_(2) hArr 2NH_(3)` `{:("Initial concentration ",0.25,0.4,0),("At equilibrium",0.25-x,0.4-3x,2x):}` Concentration of `H_(2)` at equilibrium is given as =0.1 ` therefore 0.4-3x=0.1 implies x=0.1` Concenration of `N_(2)` at equilibrium=0.25-0.1 =0.15 Concentration of `NH_(3)` at equilibrium =`0.1xx2=0.2` When the reactiion is CARRIED with double the number of moles , concentration of `N_(2)=0.15xx2=0.3` mol `L^(-1)`and CONCENTRATIONOF `NH_(3)=0.2xx2=0.4` mol `L^(-1)`.