Respuesta :
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:
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
