Question

A mixture of 0.500 mols of NO and 0.100 mols O2 are combined in a 5 L vessel. The two react according to the reaction: 2NO (g)⇋O2(g) + N2(g)

(a) At equilibrium, [NO] = 0.0125M. What the concentrations of O2 and N2 at equilibrium?

b. What is Kc of this reaction?

c. Can you determine the value of Kp with this information? Explain

please show steps and explain

Answer 1

The given reaction is

2 NO (g) O₂ (g) + N₂ (g)

given , initial moles of NO = 0.500, Volume of Flask = 5 L , So, [NO] = (0.500/5) = 0.1 M

and , initial moles of O₂ = 0.100 , hence, [O₂] = (0.100/5) = 0.02 M

now the ICE table is

	[NO2]	[O2]	[N2]
initial	0.1 M	0.02 M	0
change	(0.0125 - 0.1) = - 0.0875 M	+ (0.875/2) = + 0.04375 M	+ (0.0875/2) = 0.04375 M
Equilibrium	0.0125 M	0.02+ 0.04375 = 0.06375 M	0.04375 M

now, Kc = =  

Or , K_c = 17.85

B)

2 NO (g) O₂ (g) + N₂ (g)

we know,

K_p =K_c (RT)

where, = (moles of gaseous products - moles of gaseous reactants)

for the above reaction,

= (1+1) - 2 = 0

hence, K_p =K_c (RT)⁰

or, K_p = K_c = 17.85