Unit 7 · Topic 7.6 Beta

Properties of the Equilibrium Constant

K belongs to one equation written one way.

Practice 5: Mathematical Routines

Question set for this topic

Part 1 · Hook

Why this matters

Some reactions are hard to study directly: they are too slow, or one product reacts further. Chemists get around this the way a navigator plots a route through waypoints. They measure K for simpler reactions that add up to the one they want, then combine those K values with three short rules.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is Kc for 2 SO2(g) + O2(g) ⇌ 2 SO3(g)?

  1. [SO3]2 / ([SO2]2[O2])
  2. 2[SO3] / (2[SO2][O2])
  3. [SO2]2[O2] / [SO3]2
  4. [SO3] / ([SO2][O2])
Show the answer

Products over reactants, coefficients as exponents.

  • Correct: [SO3]2 / ([SO2]2[O2]):
  • 2[SO3] / (2[SO2][O2]):
  • [SO2]2[O2] / [SO3]2:
  • [SO3] / ([SO2][O2]):

2. In Hess’s law, reversing a reaction does what to ΔH?

  1. Changes its sign
  2. Takes its reciprocal
  3. Doubles it
  4. Leaves it unchanged
Show the answer

The reverse reaction has the same size of ΔH with the opposite sign.

  • Correct: Changes its sign:
  • Takes its reciprocal:
  • Doubles it:
  • Leaves it unchanged:

3. What does K ≪ 1 tell you?

  1. The equilibrium mixture is mostly reactants
  2. The reaction is slow
  3. The equilibrium mixture is mostly products
  4. The reaction does not occur
Show the answer

Small K: reactant-favored, with a little product.

  • Correct: The equilibrium mixture is mostly reactants:
  • The reaction is slow:
  • The equilibrium mixture is mostly products:
  • The reaction does not occur:

Part 4 · See it

See it first

Three boxes. Reverse it: A + B forming C has K; C forming A + B has 1 over K. Multiply by n: doubling to 2A + 2B forming 2C gives K to the n, here K squared. Add reactions: A forming B with K1 and B forming C with K2 add to A forming C with K1 times K2. A note: coefficients are exponents in K, so scaling raises K to a power and adding multiplies K values; for one equation, only temperature changes K.
Reverse an equation: 1/K. Multiply its coefficients by n: Kⁿ. Add equations: multiply the K values. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. An equation is reversedits K expression turns upside down, so K becomes 1/K
  2. Every coefficient is multiplied by nevery exponent in K is multiplied by n, so K becomes Kⁿ
  3. Two equations are addedmultiplying their expressions cancels the shared species, so K = K₁ × K₂
  4. The same equation is studied at a new temperatureK takes a new value; at a fixed temperature nothing else changes it

Part 6 · Key ideas

Key ideas

  • Reverse an equation: Knew = 1/K.
  • Multiply all coefficients by n: Knew = Kn (n = ½ means a square root).
  • Add equations: multiply their K values.
  • For a given equation, K depends only on temperature. Concentration, pressure, volume and catalysts do not change it.

Part 7 · Misconception

A common mistake

The wrong idea: Adding more reactant to an equilibrium mixture increases K.

What actually happens: Adding reactant changes Q, not K. The system reacts until Q equals the same K again. For a given equation, only a change of temperature changes K.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

Two reactions of nitrogen oxides

A student is given Kc values for two reactions at 25 °C and wants Kc for related reactions at 25 °C.

Given reactions
ReactionEquationKc
AN2(g) + O2(g) ⇌ 2 NO(g)4.5 × 10−31
B2 NO(g) + O2(g) ⇌ 2 NO2(g)6.0 × 1013

1. Calculate Kc for 2 NO2(g) ⇌ 2 NO(g) + O2(g).

Type a number.

Show the answer

This is reaction B reversed, so K = 1/KB = 1 / (6.0 × 10¹³) = 1.7 × 10−14.

  • Answer: 1.7 × 10-14

2. Calculate Kc for NO(g) + ½ O2(g) ⇌ NO2(g).

Type a number.

Show the answer

Reaction B with every coefficient halved: K = (KB)1/2 = √(6.0 × 10¹³) = 7.7 × 106.

  • Answer: 7.7 × 106

3. Reactions A and B can be combined to describe nitrogen dioxide forming directly from the elements. Use the table to calculate Kc at 25 °C for N2(g) + 2 O2(g) ⇌ 2 NO2(g).

Type a number.

Show the answer

Add reaction A to reaction B: 2 NO cancels, leaving N₂ + 2 O₂ ⇌ 2 NO₂. K = KA × KB = (4.5 × 10⁻³¹)(6.0 × 10¹³) = 2.7 × 10−17.

  • Answer: 2.7 × 10-17

4. Which operation gives Kc for 4 NO(g) ⇌ 2 N2(g) + 2 O2(g)?

  1. (1/KA)2
  2. 2/(KA)
  3. (KA)2
  4. −2KA
Show the answer

Reversing gives the reciprocal; multiplying all coefficients by n raises K to the n. Together: (1/KA)².

  • Correct: (1/KA)2: Right: reverse reaction A (1/KA), then double every coefficient (square it).
  • 2/(KA): Doubling the equation squares K; it does not multiply K by 2.
  • (KA)2: This doubles reaction A but does not reverse it. The new reaction has NO as the reactant.
  • −2KA: Neither reversing nor doubling makes K negative or multiplies it by 2.

Data table

K at different temperatures

Kc for N2O4(g) ⇌ 2 NO2(g) is measured at four temperatures.

Kc for N2O4(g) ⇌ 2 NO2(g)
Temperature (K)Kc
2984.6 × 10−3
3500.13
4001.5
45010.

5. A flask at 350 K holds an equilibrium mixture. More N2O4 is injected and the flask is held at 350 K. What is Kc once equilibrium is restored?

  1. 0.13
  2. Greater than 0.13, because more N2O4 reacts.
  3. Less than 0.13, because there is more reactant on the bottom of the expression.
  4. 1.5, because the system shifts to the next K in the table.
Show the answer

At a fixed temperature, K is a constant. Adding N₂O₄ lowers Q below K, and the net forward reaction restores Q = K = 0.13.

  • Correct: 0.13: Right: K depends only on temperature. Adding a reactant changes Q and the concentrations, not K.
  • Greater than 0.13, because more N2O4 reacts.: More N₂O₄ reacts, but the new concentrations still satisfy the same K. K itself does not move.
  • Less than 0.13, because there is more reactant on the bottom of the expression.: Right after the injection, Q is below 0.13. The system then reacts forward until Q is back at 0.13.
  • 1.5, because the system shifts to the next K in the table.: K changes only if the temperature changes. Here it stays at 350 K.

6. Given: (1) A(g) ⇌ B(g), K1 = 1.4 × 10−2; (2) B(g) ⇌ 2 C(g), K2 = 6.0 × 10−5. Calculate K for A(g) ⇌ 2 C(g).

Type a number.

Show the answer

Adding (1) and (2) cancels B: A ⇌ 2 C. K = K₁K₂ = (1.4 × 10⁻²)(6.0 × 10⁻⁵) = 8.4 × 10−7.

  • Answer: 8.4 × 10-7

7. Which changes alter the value of Kc for a given reaction equation? Select all that apply.

  1. Changing the temperature
  2. Adding more of a reactant
  3. Changing the volume of the container
  4. Adding a catalyst
Show the answer

For a given equation, K changes only with temperature. Writing the equation differently (reversed, scaled) gives a different K, but that is a different equation.

  • Correct: Changing the temperature: Right: K depends on temperature.
  • Adding more of a reactant: This changes Q and the concentrations, not K.
  • Changing the volume of the container: Volume changes concentrations (and Q), not K.
  • Adding a catalyst: A catalyst speeds both directions equally and leaves K unchanged.

Part 9 · Summary

Summary

K belongs to one equation written one way. Reversing the equation gives 1/K, multiplying every coefficient by n gives Kⁿ, and adding equations multiplies their K values, because K is built from products of concentration terms. For a given equation, K changes only with temperature: adding substances, changing the volume or adding a catalyst changes Q or the rates, never K.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections