Question

HCl (g) + NaOH (aq) <-> NaCl (aq) + H2O (l). What is the Q value for the equation and which direction will the reaction shift given that K = 0.5? , [HCl] = 2.5 M, [NaOH] = 3.5 M, [NaCl] = 1.75 M, [H2O] = 1 M

Answer 1

Answer:

Qc <Kc and the system will evolve to the right.

Explanation:

The reaction constant Qc is calculated when a reaction that has not yet reached equilibrium and allows determining where the reaction will move to reach equilibrium. For the following reaction:

aA + bB ⇔ cC + dD

the constant Qc is calculated as:

$Qc=\frac{[C]^{c}*[D]^{d} }{[A]^{a} *[B]^{b} }$

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

In this case:

$Qc=\frac{[NaCl]*[H_{2}O] }{[HCl]*[NaOH]}$

Being:

• [NaCl] = 1.75 M
• [H₂O] = 1 M
• [HCl] = 2.5 M
• [NaOH] = 3.5 M

Replacing:

$Qc=\frac{1.75 M*1 M}{2.5 M*3.5 M}$

Solving:

Qc=0.2

Being Kc = 0.5, then Qc <Kc and the system will evolve to the right.