Question

Having (2SO_3(g) rightleftharpoons 2SO_2(g) + O_2(g)) the (K_c = 3.57 × 10^{-3}) at (726.85^˚ {C}). What is the equilibrium (K) for the following reaction at (726.85^˚ {C}) and (0.25 , {mol}) (SO_3(g) rightleftharpoons 0.25 , {mol}) (SO_2(g) + 0.125 , {mol} O_2(g))?

- A) (0.00357)
- B) (2.64 × 10^{-20})
- C) (4.46 × 10^{-4})
- *D) (0.494)*
- E) (0.244)

Answer 1

The new equilibrium constant for the modified reaction at the same temperature is found by raising the original equilibrium constant to the power of ¼. After performing the calculation, the correct answer is D) 0.494.

Step-by-step explanation:

The student's question is asking to calculate the equilibrium constant (K) for a modified version of a given chemical reaction: SO3(g) ⇌ SO2(g) + O2(g), with the equilibrium constant at a specific temperature already provided (Kc = 3.57 × 10-3 for the reaction 2SO3(g) ⇌ 2SO2(g) + O2(g)). The modified reaction involves fractional coefficients: 0.25 SO3(g) ⇌ 0.25 SO2(g) + 0.125 O2(g).

To find the new equilibrium constant, we must consider that the expression for the equilibrium constant depends on the stoichiometry of the reaction. Since the coefficients of the species in the balanced equation have been divided by 4, we need to raise the original equilibrium constant to the power of ¼ (which is the inverse of 4). The relationship between the original Kc and the new K is given by K = (Kc)¼. Calculating this we get:

K = (3.57 × 10-3)¼
K ≈ 0.494
Therefore, the correct answer to the student's question is D) 0.494.