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High School Chemistry : Help with Reaction Quotient

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Example Question #1 : Help With Reaction Quotient

Consider the following balanced reaction:

$MgF_{2}_{(s)} \rightarrow Mg^{2+} + 2F^{-}$ $K_{sp} = 5.2*10^{-11}$

The solution has the following concentrations:

$[Mg^{2+}] = 1*10^{-4}M$

$[F^{-}] = 2*10^{-3}M$

Based on these concentrations, what will happen to the solution?

Possible Answers:

It will shift to the right

More information is needed in order to answer the question

It will shift to the left

The system is at equilibrium

Correct answer:

It will shift to the left

Explanation:

By placing these initial concentrations in the equilibrium expression, we can compare the reaction quotient, $\small Q$ , to the equilibrium constant, $\small K_{eq}$ .

The reaction quotient for this reaction is:

$[Mg^{2+}][F^{-}]^{2} = Q$

Note that the reactant is a pure solid, and will not be included in this expression. By setting this equal to the equilibrium constant, we can see whether $\small Q$ or $\small K_{eq}$ is larger in value.

$Q = [Mg^{2+}][F^{-}]^{2} = [1*10^{-4}][2*10^{-3}]^{2} = 4*10^{-10}$

We can predict how a solution will change based on the value of $\small Q$ . When $\small Q$ is less than $\small K_{eq}$ , the reaction will shift to the right. If $\small Q$ is greater than $\small K_{eq}$ , the reaction will shift to the left. If $\small Q$ is equal to $\small K_{eq}$ , the solution is at equilibrium.

$Q=4*10^{-10}$

$K_{sp}=K_{eq}=5.2*10^{-11}$

Since the value for $\small Q$ is greater than the equilibrium constant in this instance, the reaction will shift to the left.

$4*10^{-10}>5.2*10^{-11}$

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Example Question #56 : Reactions And Equilibrium

Consider the following reaction.

$H_3(PO_4)_2(aq)\rightleftharpoons 3H^++2PO_4^{3-}$

What is the reaction quotient if $[H_3(PO_4)_2]=[H^+]=[PO_4^{3-}]=1.00\hspace{1 mm}M$ ?

Possible Answers:

1.00

0.167

There is not enough information to answer this question

6.00

1.677

Correct answer:

1.00

Explanation:

$H_3(PO_4)_2\rightleftharpoons 3H^++2PO_4^{3-}$

The reaction quotient, or Q, of the above reaction is equal to the products over the reactants. Q is calculated in the same manner as K_eq, but does not require that the reaction be at equilibrium.

$Q=\frac{[products]^a}{[reactants]^b}$

$\frac{[H^+]^3[PO_4^{3-}]^2}{[H_3(PO_4)_2]}=\frac{[1.00]^3[1.00]^2}{[1.00]}=1.00$

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High School Chemistry : Help with Reaction Quotient

Study concepts, example questions & explanations for High School Chemistry

All High School Chemistry Resources

Example Questions

Example Question #1 : Help With Reaction Quotient

Example Question #56 : Reactions And Equilibrium

All High School Chemistry Resources

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