At 585 K and a total pressure of 1.00 atm, NOCl(g) is 56.4/0 dissociated. Assume that 1.00
mol of ONCl(g) was present before dissociation, (a) How many moles of ONCl(g), NO(g),
and Cl2(g) are present at equilibrium? (b) What is the total number of moles of gas present
at equilibrium? (c) What are the equilibrium partial pressures of the three gases? (d) What
is the numerical value of Kp at 585 K?
To answer these questions, we will use the concept of equilibrium and the given information about the extent of dissociation of NOCl(g).
(a) To find the number of moles of ONCl(g), NO(g), and Cl2(g) present at equilibrium, we need to determine the moles of each species that dissociate. In this case, it is given that NOCl(g) is 56.4% dissociated.
Given that 1.00 mol of ONCl(g) was initially present:
- The moles of ONCl(g) that dissociate = 1.00 mol x 0.564 = 0.564 mol
- The moles of ONCl(g) remaining at equilibrium = 1.00 mol - 0.564 mol = 0.436 mol
Since ONCl(g) dissociates into NO(g) and Cl2(g), the moles of NO(g) and Cl2(g) can be assumed equal to the moles of ONCl(g) dissociated:
- Moles of NO(g) = 0.564 mol
- Moles of Cl2(g) = 0.564 mol
(b) The total number of moles of gas present at equilibrium is the sum of the moles of ONCl(g), NO(g), and Cl2(g):
- Total moles of gas = 0.436 mol + 0.564 mol + 0.564 mol = 1.564 mol
(c) To find the equilibrium partial pressures, we need to consider the ideal gas law and the concept of mole fraction.
The mole fraction (x) of a component in a mixture is given by the ratio of its moles to the total moles of the mixture. The partial pressure (P) of a gas in the mixture is related to its mole fraction:
P = x * P_total
The mole fraction of ONCl(g) is:
- Mole fraction of ONCl(g) = (moles of ONCl(g) / total moles of gas) = 0.436 mol / 1.564 mol = 0.2789
Similarly, the mole fractions of NO(g) and Cl2(g) are also 0.2789 since their moles are equal to the moles of ONCl(g) dissociated.
To find the equilibrium partial pressures, multiply the mole fraction of each gas by the total pressure (1.00 atm):
- Partial pressure of ONCl(g) = x * P_total = 0.2789 * 1.00 atm = 0.2789 atm
- Partial pressure of NO(g) = x * P_total = 0.2789 * 1.00 atm = 0.2789 atm
- Partial pressure of Cl2(g) = x * P_total = 0.2789 * 1.00 atm = 0.2789 atm
(d) The numerical value of Kp at 585 K can be determined using the equilibrium concentrations (or partial pressures) of the reactants and products.
In this case, Kp is the equilibrium constant expressed in terms of partial pressures. Since the reaction is:
2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)
The equilibrium constant expression for the reaction is:
Kp = (P_NO)^2 * P_Cl2 / (P_NOCl)^2
Substituting the given partial pressures into the equation, we get:
Kp = (0.2789 atm)^2 * (0.2789 atm) / (0.2789 atm)^2
Simplifying:
Kp = 0.078
Therefore, at 585 K, the numerical value of Kp is 0.078.