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.

Question

Georgia Sheppard

Chemistry

8 September, 20:12

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.

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Answers (2)

Know the Answer?

Pasha · Answer 1

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

Keshawn Moore · Answer 2

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