Consider the equilibrium 2NO (g) <---> 2NO (g) + Cl2 (g). In a 1 L container @ equilibrium there are 1.0 mol NOCL, 0.70 mol NO, and 0.40 mol Cl2. @ constant temperature and volume, 0.10 mol NaCl is added. What are the concentrations in the "new" equilibrium in comparison to the concentrations in the "old" equilibrium?
You may want to clarify the problem. Is that 2NOCl ==> 2NO + Cl2?
Is that NaCl or NOCl added? Adding NaCl shouln't change anything.
To determine the concentrations in the "new" equilibrium, we can use the concept of the reaction quotient (Q). The reaction quotient helps us compare the concentrations of the reactants and products at any given time during a reaction. For a reaction in the form aA + bB ⇌ cC + dD, the reaction quotient (Q) is given by:
Q = [C]^c[D]^d/[A]^a[B]^b
In our case, the reaction is 2NO(g) ⇌ 2NOCL(g) + Cl2(g), so the reaction quotient (Q) would be:
Q = ([NOCL]^2[Cl2])/([NO]^2)
In the "old" equilibrium, the concentrations are: [NOCL] = 1.0 mol, [NO] = 0.70 mol, and [Cl2] = 0.40 mol.
Substituting these values into the reaction quotient (Q):
Q(old) = (1.0^2 * 0.40) / (0.70^2) = 0.8163 (approximately)
Now, let's consider the addition of 0.10 mol NaCl. NaCl is not part of the equilibrium reaction, so adding it will not affect the Q value directly. However, the addition of NaCl will affect the concentrations of the reactants and therefore the ratio of the concentrations.
To determine the concentrations in the "new" equilibrium, we need to consider that one mole of NaCl will react with one mole of NOCL according to the equation:
NaCl + NOCL ⇌ NO + Cl2
Since 0.10 mol of NaCl is added, it will react with 0.10 mol of NOCL. As a result, the concentration of NOCL will decrease by 0.10 mol, and the concentrations of NO and Cl2 will increase by 0.10 mol each.
In the "new" equilibrium, the concentrations will be:
[NOCL] = 1.0 mol - 0.10 mol = 0.90 mol
[NO] = 0.70 mol + 0.10 mol = 0.80 mol
[Cl2] = 0.40 mol + 0.10 mol = 0.50 mol
Now we can calculate the Q value for the "new" equilibrium:
Q(new) = (0.90^2 * 0.50) / (0.80^2) ≈ 0.844
Comparing Q(new) to Q(old), we see that the Q value has increased slightly. This means that the concentrations in the "new" equilibrium, particularly the concentrations of NOCL and Cl2, have increased compared to the "old" equilibrium. However, the concentration of NO has decreased compared to the "old" equilibrium.