5.6 x 10-6 mol of A and 5 x 10-5 mol of B are mixed in a 200 mL flask. The system is represented by the equation:
2A(g) + B(g) --> 3C(g)
At equilibrium, there is 4.8 x 10-5 mol of B. Calculate the value of the equilibrium constant.
To calculate the value of the equilibrium constant, also known as the equilibrium constant expression (Kc), we need to use the concentrations of reactants and products at equilibrium.
Given information:
Initial moles of A = 5.6 x 10^(-6) mol
Initial moles of B = 5 x 10^(-5) mol
Moles of B at equilibrium = 4.8 x 10^(-5) mol
To find the moles of A and C at equilibrium, we use the stoichiometry of the reaction. From the balanced equation, we can see that for every 2 moles of A consumed, 1 mole of B is consumed and 3 moles of C are produced.
Moles of A consumed = (initial moles of A) - [ (moles of B at equilibrium) / 2 ]
= (5.6 x 10^(-6) mol) - [ (4.8 x 10^(-5) mol) / 2 ]
= 5.6 x 10^(-6) mol - 2.4 x 10^(-5) mol
= -1.84 x 10^(-5) mol (negative sign indicates consumption)
Moles of C produced = (moles of B at equilibrium) * 3
= (4.8 x 10^(-5) mol) * 3
= 1.44 x 10^(-4) mol
Now, let's calculate the concentrations at equilibrium, knowing that the total volume is 200 mL.
Concentration of A at equilibrium (C_A) = (moles of A at equilibrium) / (total volume)
= (-1.84 x 10^(-5) mol) / (0.2 L)
= -9.2 x 10^(-5) mol/L (negative sign indicates consumption)
Concentration of B at equilibrium (C_B) = (moles of B at equilibrium) / (total volume)
= (4.8 x 10^(-5) mol) / (0.2 L)
= 2.4 x 10^(-4) mol/L
Concentration of C at equilibrium (C_C) = (moles of C at equilibrium) / (total volume)
= (1.44 x 10^(-4) mol) / (0.2 L)
= 7.2 x 10^(-4) mol/L
Now we can write the equilibrium constant expression (Kc):
Kc = (C_C)^3 / (C_A)^2 * (C_B)
= (7.2 x 10^(-4))^3 / (-9.2 x 10^(-5))^2 * (2.4 x 10^(-4))
= 419.37
Therefore, the value of the equilibrium constant (Kc) is approximately 419.37.