Gas A is in equilibrium with gases B and C according to the following equation - Leaving Cert Chemistry - Question b - 2015

To find the number of gaseous moles at equilibrium, we start with the initial concentration of gas A. We have 30 moles of gas A in a 10-litre container:

• Initial concentration of A: $C_A = \frac{30 \text{ moles}}{10 \text{ litres}} = 3.0 \text{ mol/L}$

Let the change in concentration of A at equilibrium be denoted as 'x'. The equilibrium expression is:

$K_c = \frac{[B][C]}{[A]} \Rightarrow K_c = \frac{(x)(x)}{(3.0 - x)} = 4$

This can be rearranged into:

$4(3.0 - x) = x^2 \Rightarrow 12 - 4x = x^2 \Rightarrow x^2 + 4x - 12 = 0$

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 + 48}}{2} = \frac{-4 \pm 8}{2}$

This gives us: $x = 2 ext{ or } x = -6$

Since x must be positive, we have x = 2. Thus, at equilibrium:

• Moles of gas A = 3.0 - 2 = 1.0 moles.
• Moles of gas B = 2.
• Moles of gas C = 2.

Total moles at equilibrium = 1.0 + 2 + 2 = 5.0 moles.