Question

The following equilibrium reaction is investigated.

2 N2O(g) + N2H4(g) ⇆ 3 N2(g) + 2 H2O(g)

  When 0.10 moles of N2O and 0.25 moles of N2H4 are placed in a 10.0-L container and allowed to come to equilibrium, the equilibrium concentration of N2O is 0.0060 M. What is the equilibrium concentration of all four substances?

Answer 1

initially,
[N2O] = number of mol / volume
= 0.10 mol / 10 L
=0.01 M

[N2H4] = number of mol / volume
= 0.25 mol / 10 L
=0.025 M

2 N2O(g) + N2H4(g) <-----> 3 N2(g) + 2 H2O(g)
0.01                 0.025                      0                  0              (initial)
0.01-2x           0.025-x                  3x               2x             (at equilibrium)

given at equilibrium,
[N2O] = 0.01-2x = 0.0060
x = 0.002 M

equilibrium concentration will be:
[N2O] = 0.0060 M
[N2H4] = 0.025 - x = 0.025 - 0.0020 = 0.023 M
[N2] = 3x = 3*0.002 M = 0.006 M
[H2O] = 2x = 2*0.002 M = 0.004 M