Question

1. A mixture of 0.200 moles of N₂O and 0.100 moles of O₂ is placed in a reaction container and allowed to react until equilibrium is established.

2 N₂O(g) + O₂(g) ↔ 4 NO(g)

At equilibrium, 0.196 moles of NO is present. What is the composition of the equilibrium mixture in terms of moles of each substance present?

2. A 0.178 mole sample of C₂H₆ and a 0.329 moles sample of H₂ are placed in a reaction container and allowed to react until equilibrium is established.

C₂H₆(g) + H₂(g) ↔ 2 CH₄(g)

At equilibrium, it is found that 0.010 mole of H₂ has reacted. What is the composition of the equilibrium mixture in terms of moles of each substance present?

Answer 1

a)

2N₂O + O₂ <----> 4NO

	N2O	O2	NO
Initial	0.200 mol	0.100 mol	0 mol
Change	-2x	-x	+4x
Equilibrium	(0.200-2x) mol	(0.100-x) mol	4x

Now,

4x = 0.196 moles

x = 0.049 mol

At equilibrium,

[N₂O] = 0.200 - 2(0.049) = 0.102 mol

[O₂] = 0.100 - 0.049 = 0.051 mol

[NO] = 0.196 mol

b)

C₂H₆ + H₂ <------> 2CH₄

	C2H6	H2	CH4
Initial	0.178 mol	0.329 mol	0 mol
Change	-x	-x	+2x
Equilibrium	(0.178-x) mol	(0.329-x) mol	2x

Now,

x = 0.010 mol

At equilibrium,

[C₂H₆] = 0.178 - 0.010 = 0.168 mol

[H₂] = 0.329 - 0.010 = 0.319 mol

[CH₄] = 2(0.010) = 0.020 mol