Question

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures of 9.30×10−2 atm , 0.174 atm , and 0.25 atm for NO, Cl2, and NOCl, respectively. Calculate Kp for this reaction at 500 K. Then if the vessel has a volume of 5.20 L calculate Kc.

Answer 1

Answer:

Kp = 41.53

Kc = 1.01

Explanation:

To calculate the equilibrium constant in terms of pressure, what we simply do is to use the equilibrium pressure raised to the power of the number of moles. What we are saying in essence is this:

Kp = [NOCl]^2/[NO]^2[Cl]

Kp= [0.25]^2/[0.174][0.093]^2 = 41.53

Kp = Kc (RT)^Dn

Hence, Kc = Kp/[RT]^(delta n )^-1

n = sum of the number of moles of products minus the sum of the number of moles of reactants= 2-3 = -1 in this case

Kc = 41.53/(0.0821 * 500)^1

Kc = 1.01

Answer 2

Answer:

a) Kp = 4.9545 E-3

b) Kc = 0.2016

Explanation:

• 2NO(g) + Cl2(g) ↔ 2NOCl(g)

eq. mix:

∴ Pp NO = 9.30 E-2 atm

∴ Pp Cl2 = 0.174 atm

∴ Pp NOCl = 0.25 atm

a) Kp = (PNOCl/P°)² / (PCl2/P°)(PNO/P°)²

∴ P° = 1 atm

∴          ni       change       neq

NO    nNO    nNO - x     nNO - x

Cl2    nCl2    nCl2 - x    nCl2 - x

NOCl   0         0 + x            x

⇒ nNO = (PpNO)(V)/(R)(T) = (9.3 E-2)(5.2)/(0.082)(500) = 0.012 mol NO

⇒ nCl2 = (0.174)(5.2)/(0.082)(500) = 0.022 mol Cl2

equilibrium:

⇒ neq = nNO - x + nCl2 - x + x = PeqV/RT

⇒ neq = 0.012 - x + 0.022 - x + x  = PeqV/RT

∴ Peq = PpNO + PpCl2 + PpNOCl = 9.3 E-2 + 0.174 + 0.25 = 0.2662 atm

⇒ 0.0341 - x = (0.2662)(5.2)/(0.082)(500) = 0.0337 mol

⇒ x = 3.3805 E-4 mol eq

∴ Peq = neqRT/V

⇒ PNOCleq = (3.3805 E-4 mol)(0.082 atmL/Kmol)(500K)/(5.2 L) = 2.67 E-3 atm

⇒ PCl2eq = (0.022 - 3.3805 E-4)(0.082)(500)/(5.2) = 0.17 atm

⇒ PNOeq = (0.012 - 3.3805 E-4)(0.082)(500)/(5.2) = 0.092 atm

⇒ Kp = (2.67 E-3/1)²/(0.17/1)(0.092/1)² = 4.9545 E-3

b) Kc = ([NOCl]eq)² / ([Cl2]eq)([NO]eq)²

∴ [NOCl]eq = neq/V = 3.3805 E-4 mol/ 5.2 L = 6.500 E-5 M

∴ [Cl2]eq = (0.022 - 3.3805 E-4)mol/(5.2L) = 4.166 E-3 M

∴ [NO]eq = (0.012 - 3.3805 E-4)mol/(5.2 L) = 2.243 E-3 M

⇒ Kc = (6.500 E-5)²/(4.166 E-3)(2.243 E-3)² = 0.2016