Answer:
a) Kp = 4.9545 E-3
b) Kc = 0.2016
Step-by-step 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