Match the action to the effect on the equilibrium position for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g). (3 points)

Chemistry

1 answer:

AleksandrR [38]1 year ago

5 0

Answer:

1. Adding hydrogen gas, b. shift to the right
2. Adding a catalyst, c. No effect
3. Decreasing the pressure, a. shift to the left

Explanation:

Hydrogen gas can be rewritten as H2. Whenever you add something to an equilibrium expression, it will shift to whichever side does not have this. So, since the reactant side has 3 moles of H2, adding more H2 to the reaction will shift to the products side, since there is no H2 there.

Adding a catalyst has no effect on equilibrium reactions.

When decreasing the pressure, equilibrium will shift to the side with the greater number of moles of gas. In this case, there are 4 moles of gas on the left, and 2 on the right, so it would shift to the left.

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Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find

FromTheMoon [43]

Answer: The concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

Explanation:

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

$H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)$