Question

For the following reaction, how will the reaction equilibrium be affected by an increase

in volume of the container?
H₂O2(1) <--> H2(g) + O2(g) AH=+187 kJ
a. It will shift in favor of the products
b. It will shift in favor of the reactants
c. There will be no change

Answer 1

Okay, let's think through this step-by-step:

1) The initial equilibrium lies on the right side, favoring the products (H2 and O2 gases), because the standard enthalpy change (AH) is positive for this reaction, meaning the products are more stable.

2) When we increase the volume of the container, the pressure decreases according to Boyle's law (P=k/V).

3) A decrease in pressure favors the side with the greater number of moles of gases. In this case, the product side has 2 moles of gas (H2 + O2), so the equilibrium will shift to the right.

4) Therefore, when the volume increases, the equilibrium will shift further in favor of the products (H2 and O2 gases).

The answer is a: It will shift in favor of the products.

Let me know if this makes sense! I can re-explain anything that is unclear.

Answer 2

anwser is a because i took the test and that's what the correct anwser was