Question

The reaction 2NO(g) + O2(g) ? 2NO2(g) has Kp = 1.65 ×10^11 at 25°C. To a 1.00 L flask, 1.03 g of NO and 532 mL of O2 measured at 29°C and 772 torr are mixed. When the mixture comes to equilibrium at 25°C, what is the concentration of NO(g)? If needed, use “E” for scientific notation. Do not enter units as part of your answer. Report your answer to the correct number of significant figures.

Answer 1

Sol.

Reaction :

2NO(g) + O2(g) <----> 2NO2(g)   

As Initial moles of NO = initial mass of NO / molar mass of NO  

= 1.03 g / 30.01 g/mol  

= 0.0343 mol

Volume of flask = 1 L  

So , initial Concentration of NO = 0.0343 mol / 1 L = 0.0343 M

Initial Volume of O2 = 532 mL = 0.532 L

( As 1 L = 1000 mL )

Initial Temperature = 29°C = 29 + 273.15 = 302.15 K

Pressure = 772 torr = 772 / 760 = 1.0158 atm

( As 1 atm = 760 torr )

Gas constant = R = 0.0821 L atm / K mol  

So , Initial moles of O2  

= Pressure × Initial Volume of O2 / ( Gas constant × Initial Temperature )

= 1.0158 atm × 0.532 L / ( 0.0821 L atm / K mol × 302.15 K )

= 0.0218 mol  

and , Initial Concentration of O2  

= 0.0218 mol / 1 L = 0.0218 M

Now ,  ng = Change in number of gaseous molecules  

= Number of gaseous molecules at product side - Number of gaseous molecules at reactant side  

= 2 - 2 - 1 = - 1

Gas constant = R = 0.0821 L atm / K mol  

Equilibrium Temperature = T = 25°C = 25 + 273.15 = 298.15 K

Pressure Equilibrium Constant = Kp = 1.65 × 10¹¹   

So , Concentration Equilibrium Constant  

= Kc = Kp / ( RT)^ng

= 1.65 × 10¹¹ / ( 0.0821 × 298.15 )^-1

= 40.3889 × 10¹¹

Assume that the change in concentration at equilibrium be x

2NO(g) + O2(g) <----> 2NO2(g)

initial 0.0343 0.0218 0

change - 2x - x +2x

equilibrium 0.0343 - 2x 0.0218 - x 2x

As the value of Kc is so high , so , assume that all the NO reactant is consumed  

So , 0.0343 - 2x = 0  

2x = 0.0343

x = 0.0343 / 2 = 0.01715

So , Equilibrium Concentration of O2 = [O2]  

= 0.0218 - x = (0.0218 - 0.01715) M = 0.00465 M

Equilibrium Concentration of NO2 = [NO2]

= 2x = 0.0343 M  

Now , Kc = [NO2]² / ( [NO]² [O2] )

40.3889 × 10¹¹ = (0.0343)² / ( [NO]² × 0.00465 )  

[NO]² = 0.062643 × 10^-12

[NO] = 2.50 × 10^-7 M   

Therefore , The equilibrium concentration of NO is  

2.50 × 10^-7 M