Question

Consider the reaction 2 NH₃ (g) ? N₂ (g) + 3 H₂ (g). Suppose 6 moles of pure ammonia are placed in a 1.0 liter flask and allowed to reach equilibrium. If X represents the concentration (in M) of nitrogen present in the system once equilibrium is reached, which of the following will represent the concentration of ammonia in M? Show work

A) 6

Answer 1

2 NH₃ (g) N₂ (g) + 3 H₂ (g)

Accordin g to the balanced equation, 2 moles of NH₃ (g) produces 1 mole of N₂ (g)

Given initial moles of NH₃ (g) is 6 mol

                                        2 NH₃ (g) N₂ (g) + 3 H₂ (g)

initial moles                         6                  0            0

change                              -2a                +a        +3a

Equb moles                      6-2a                a            3a

Given that concentration (in M) of nitrogen present in the system once equilibrium is reached is X So a = X

The relation becomes

                                        2 NH₃ (g) N₂ (g) + 3 H₂ (g)

initial moles                         6                  0            0

change                              -2X                +X        +3X

Equb moles                      6-2X                X            3X

Therefore the Equilibrium moles of moles of ammonia, NH₃ is 6-2X

Given volume of the container where reaction takes place is 1.00 L

So Molarity of ammonia , M = number of moles at equilibrium / volume in L

                                         = ( 6-2X)mol / 1.00 L

                                         = 6.2X M

Therefore option ( C) is correct.