Answer to Question #171491 in General Chemistry for Elena

Question

in a container with a volume of 2 liters are introduced 4 moles of NO and 2 moles of NO2 which establish the balance:

2NO (g) + O2 (g) → 2 NO2 (g)

after spending 10% of one of the substances.

—Find Kp and Kc at temperatures 127•C .

—How pressure affects the formation of 2NO₂

—for this reaction ΔH = -114 kJ, with increasing temperature will the equilibrium constant increase or decrease?

Answer 1

I) moles of NO= 4moles

Moles of NO2= 2moles

But from the stoichiometric equation, the number of moles of these substances that react will be

Moles of NO= 2mol

Moles of NO2= 2mol (it is the limiting reagent)

Moles of O2= 1mol

After spending 10%

Molarity of NO= 90% of mole/litres = 90/100 x 2/2 = 0.90M

Molarity of NO2= 90/100 X 2/2= 0.90M

Molarity of O2= 90/100 x 1/2= 0.45M

Kc= [NO2]²/[NO]² [O2]

Kc=[0.90]²/[0.90]² [0.45]

K = 2.22

Now,

T= 127°C= 400K

Kp= Kc(RT)^∆n

But ∆n= moles of Gaseous products - moles of Gaseous reactants

= 2-3=-1

Kp= Kc(RT)-¹

Kp= 2.22(8.314 x 400)-¹

Kp= 6.68 x 10-⁴

I) increase in pressure favours the formation of NO2 and vice versa.

III) the equilibrium constant will decrease with increase in temperature for an exothermic process.