Related Problems with Solution
Problem 3 : For the reaction:
2A(g) + B(g) ⇌ 3C(g) Kc = 5.0 x 10⁻³
Given:
- Initial concentration of A = 0.20 M
- Initial concentration of B = 0.10 M
Calculate the equilibrium concentrations of A, B, and C.
Solution :
Let x be the change in concentration at equilibrium: [A] = 0.20 - 2x [B] = 0.10 - x [C] = 3x
Using the equilibrium constant expression: Kc = ([C]³) / ([A]²[B]) 5.0 x 10⁻³ = (3x)³ / ((0.20 - 2x)²(0.10 - x))
Now, we can solve for x. Since Kc is relatively small, we can make an approximation that x will be small compared to the initial concentrations:
5.0 x 10⁻³ ≈ (27x³) / ((0.20)(0.10)²)
27x³ ≈ (5.0 x 10⁻³) * 0.0002
27x³ ≈ 1.0 x 10⁻⁴
x³ ≈ (1.0 x 10⁻⁴) / 27
x³ ≈ 3.7 x 10⁻⁶
x ≈ ∛(3.7 x 10⁻⁶) ≈ 0.0151
So, the equilibrium concentrations are: [A] ≈ 0.20 - 2(0.0151) ≈ 0.1698 M [B] ≈ 0.10 - 0.0151 ≈ 0.0849 M [C] ≈ 3(0.0151) ≈ 0.0453 M
