Write the K_c expression for 2SO_2(g) + O_2(g) 2SO_3(g). [2 points]

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How are the reaction quotient and the equilibrium constant defined, and how do they predict the direction of reaction?

Topic 7.3 Reaction Quotient and Equilibrium Constant: write the expression for the reaction quotient Q and the equilibrium constant K, and compare Q with K to predict the direction of reaction.

A focused answer to AP Chemistry Topic 7.3, covering the reaction quotient Q, the equilibrium constant K, the law of mass action, Kc and Kp, and comparing Q with K to predict the direction a reaction will shift, with full worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

For a reaction $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$ , the equilibrium constant is $K = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$ , the products over the reactants, each raised to its coefficient, using equilibrium concentrations. The reaction quotient $Q$ has the same form but uses the concentrations at any moment, not just at equilibrium. Comparing them predicts the direction: if $Q < K$ there is too little product, so the reaction shifts toward products; if $Q > K$ there is too much product, so it shifts toward reactants; if $Q = K$ the system is at equilibrium. Use $K_c$ with concentrations and $K_p$ with partial pressures. Pure solids and liquids are left out of the expression because their concentrations do not change.

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What this topic is asking
The equilibrium constant expression
The reaction quotient
Comparing Q with K to predict direction
Try this

What this topic is asking

The College Board (Topic 7.3) wants you to write the expression for the reaction quotient $Q$ and the equilibrium constant $K$ , and to compare $Q$ with $K$ to predict the direction a reaction will shift. This is the quantitative heart of equilibrium: a single number that says how far from equilibrium a system is and which way it must move.

The equilibrium constant expression

This is the law of mass action. Pure solids and pure liquids are omitted because their effective concentration (activity) is constant and does not affect the position of equilibrium. The value of $K$ depends only on temperature for a given reaction.

The reaction quotient

So $Q$ is a snapshot and $K$ is the target. At equilibrium $Q = K$ . Away from equilibrium, $Q$ differs from $K$ , and the size and sign of the difference tell you how far off and in which direction.

Comparing Q with K to predict direction

The comparison gives a clean rule:

Q < K \Rightarrow \text{shift toward products}; \qquad Q > K \Rightarrow \text{shift toward reactants}; \qquad Q = K \Rightarrow \text{at equilibrium}

If $Q < K$ , the numerator (products) is too small, so the forward reaction runs to raise it until $Q = K$ . If $Q > K$ , there is too much product, so the reverse reaction runs to lower the numerator. The system always moves to make $Q$ approach $K$ . This single comparison underlies all the qualitative predictions of Le Chatelier's principle later in the unit.

A useful way to picture it is that $Q$ chases $K$ . Whatever the starting mixture, the reaction proceeds in the direction that brings $Q$ closer to the fixed target $K$ , and it stops only when they are equal. Because $K$ depends only on temperature, the target does not move unless the temperature changes; every other disturbance simply displaces $Q$ , and the system responds by shifting back. This is why the comparison of $Q$ with $K$ is the single most useful calculation in the whole equilibrium unit: it predicts the direction of reaction from any set of concentrations, whether the system is starting fresh, has just been disturbed, or is being checked to see whether it has reached equilibrium at all.

Predicting direction from Q and K

For $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ , $K_c = 50.$ at a certain temperature. A mixture has $[\text{H}_2] = 0.20$ , $[\text{I}_2] = 0.20$ , $[\text{HI}] = 1.0$ M.

step 1 Write the Q expression

$Q_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$ .

step 2 Substitute the current concentrations

$Q_c = \dfrac{(1.0)^2}{(0.20)(0.20)} = \dfrac{1.0}{0.040} = 25$ .

step 3 Compare Q with K

$Q_c = 25 < K_c = 50.$

step 4 Predict the direction

Since $Q < K$ , the reaction shifts toward products (more HI forms) until $Q$ rises to equal $K$ .

Try this

Q1. Write the $K_c$ expression for $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ . [2 points]

Cue. $K_c = \dfrac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}$ .

Q2. A system has $Q = 8.0$ and $K = 8.0$ . State what this tells you. [1 point]

Cue. $Q = K$ , so the system is at equilibrium and there is no net shift.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). For

\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)

K_c = 0.50

at a certain temperature. A mixture has

[\text{N}_2] = 1.0

[\text{H}_2] = 1.0

[\text{NH}_3] = 1.0

M. (a) Write the expression for

K_c

. (b) Calculate the reaction quotient

Q_c

for this mixture. (c) Compare

Q_c

with

K_c

to predict the direction the reaction will shift. (d) Justify your prediction in terms of

Q

and

K

Show worked answer →

A 4-point quantitative FRQ on Q and K.

(a) Expression (1 point): $K_c = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$ .
(b) Reaction quotient (1 point): $Q_c = \dfrac{(1.0)^2}{(1.0)(1.0)^3} = 1.0$ .
(c) Direction (1 point): $Q_c = 1.0 > K_c = 0.50$ , so the reaction shifts to the left (toward reactants).
(d) Justify (1 point): when $Q > K$ there is too much product relative to equilibrium, so the reverse reaction is favored and the system shifts left until $Q$ falls to $K$ .

Markers reward the $K_c$ expression, the value of $Q_c$ , the leftward shift, and the reasoning that $Q > K$ favors the reverse reaction.

AP 2021 (style)1 marksSection I (multiple choice). For a reaction at a given moment,

Q < K

. The reaction will (A) shift toward products (B) shift toward reactants (C) already be at equilibrium (D) stop entirely. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (A).

When $Q < K$ there is too little product relative to equilibrium, so the forward reaction is favored and the system shifts toward products until $Q$ rises to equal $K$ . The trap is reversing the logic: $Q < K$ favors products, $Q > K$ favors reactants.

Related dot points

Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)