College Chemistry Quiz: Calculating Equilibrium Concentrations

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Calculating Equilibrium ConcentrationsQuestion 1 of 11

For the reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ , $K_c = 0.36$ at 100°C. A mixture initially contains 0.200 M $N_2O_4$ and 0.100 M $NO_2$ . What is the equilibrium concentration of $N_2O_4$ ?

0.085 M

0.105 M

0.125 M

0.145 M

0.165 M

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College Chemistry Quiz

College Chemistry Quiz: Calculating Equilibrium Concentrations

Practice Calculating Equilibrium Concentrations in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Calculating Equilibrium Concentrations, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

0.085 M
0.105 M
0.125 M (correct answer)
0.145 M
0.165 M

Explanation: When you encounter an equilibrium problem with initial concentrations that aren't at equilibrium, you need to determine which direction the reaction will proceed and then calculate the final equilibrium concentrations using an ICE table. First, calculate the reaction quotient

Q_c

using the initial concentrations:

Q_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(0.100)^2}{0.200} = 0.050

. Since

Q_c = 0.050 < K_c = 0.36

, the reaction must shift right to reach equilibrium. Set up an ICE table. Let

x

be the amount of

N_2O_4

that decomposes:

Initial: $[N_2O_4] = 0.200$ , $[NO_2] = 0.100$
Change: $[N_2O_4] = -x$ , $[NO_2] = +2x$
Equilibrium: $[N_2O_4] = 0.200-x$ , $[NO_2] = 0.100+2x$

Substitute into the equilibrium expression:

K_c = \frac{(0.100+2x)^2}{0.200-x} = 0.36

Expanding:

(0.100+2x)^2 = 0.36(0.200-x)

0.010 + 0.400x + 4x^2 = 0.072 - 0.36x

4x^2 + 0.760x - 0.062 = 0

Using the quadratic formula:

x = 0.075

M Therefore:

[N_2O_4] = 0.200 - 0.075 = 0.125

M Answer C (0.125 M) is correct. Answer A (0.085 M) likely comes from calculation errors in the quadratic formula. Answer B (0.105 M) might result from assuming the wrong direction of shift. Answer D (0.145 M) could stem from sign errors in the ICE table setup. Always check that

Q_c

K_c

to determine reaction direction before setting up your ICE table—this prevents costly directional errors.

Question 2

The equilibrium $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$ has $K_p = 0.020$ at 500°C. If the equilibrium pressures are $P_{H_2} = 0.15$ atm and $P_{I_2} = 0.25$ atm, what is the equilibrium pressure of $HI$ ?

1.2 atm
1.4 atm (correct answer)
1.6 atm
1.8 atm
2.0 atm

Explanation: When you encounter equilibrium problems involving gas-phase reactions, you need to use the equilibrium constant expression

K_p

to relate the partial pressures of all species at equilibrium. For the reaction

2HI(g) \rightleftharpoons H_2(g) + I_2(g)

, the equilibrium expression is:

K_p = \frac{P_{H_2} \times P_{I_2}}{P_{HI}^2} = 0.020

Notice that

P_{HI}

is squared because the stoichiometric coefficient of

HI

is 2 in the balanced equation. Substituting the given values:

0.020 = \frac{(0.15)(0.25)}{P_{HI}^2} = \frac{0.0375}{P_{HI}^2}

Solving for

P_{HI}

P_{HI}^2 = \frac{0.0375}{0.020} = 1.875

P_{HI} = \sqrt{1.875} = 1.37 \text{ atm}

This rounds to 1.4 atm, confirming answer B. Answer A (1.2 atm) would result if you incorrectly calculated

\sqrt{1.875}

or made an arithmetic error. Answer C (1.6 atm) might come from forgetting to square the

HI

pressure in the denominator. Answer D (1.8 atm) could result from using the wrong equilibrium expression or calculation errors. Remember to always write the

K_p

expression carefully, paying attention to stoichiometric coefficients that become exponents. Double-check your algebra when solving for unknown pressures, especially when square roots are involved.

Question 3

At 25°C, the equilibrium $Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons FeSCN^{2+}(aq)$ has $K_c = 890$ . If a solution initially contains 0.00200 M $Fe^{3+}$ and 0.00150 M $SCN^-$ , what is the equilibrium concentration of $FeSCN^{2+}$ ?

0.00135 M
0.00145 M (correct answer)
0.00155 M
0.00165 M
0.00175 M

Explanation: When you encounter equilibrium problems with large equilibrium constants, you need to set up an ICE table and solve systematically, but be prepared that the equilibrium might strongly favor products. Set up your ICE table for this reaction. Initially, you have 0.00200 M

Fe^{3+}

and 0.00150 M

SCN^-

, with zero

FeSCN^{2+}

. Let

x

represent the amount of

FeSCN^{2+}

formed at equilibrium. Since

SCN^-

is the limiting reagent (smaller initial concentration), the maximum possible

x

is 0.00150 M. At equilibrium:

[Fe^{3+}] = 0.00200 - x

[SCN^-] = 0.00150 - x

, and

[FeSCN^{2+}] = x

. The equilibrium expression is:

K_c = \frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^-]} = 890

Substituting:

890 = \frac{x}{(0.00200-x)(0.00150-x)}

Since

K_c

is large, most reactants convert to products. Expanding and rearranging gives:

890x^2 - 3.115x + 0.00267 = 0

Using the quadratic formula:

x = 0.00145

M (the physically meaningful root). Answer choice A (0.00135 M) underestimates the extent of reaction given the large

K_c

value. Answer choice C (0.00155 M) exceeds what's possible given the limiting reagent constraint. Answer choice D (0.00165 M) is impossible since it exceeds the initial

SCN^-

concentration. Study tip: For large

K_c

values (>100), expect nearly complete conversion of the limiting reagent, but always solve systematically rather than assuming 100% conversion.

Question 4

For $2NO_2(g) \rightleftharpoons N_2O_4(g)$ , $K_c = 170$ at 25°C. A reaction mixture initially contains 0.040 M $NO_2$ and 0.010 M $N_2O_4$ . What is the equilibrium concentration of $NO_2$ ?

0.012 M
0.016 M (correct answer)
0.020 M
0.024 M
0.028 M

Explanation: When you encounter an equilibrium problem with initial concentrations that aren't at equilibrium, you need to determine the reaction direction and calculate how concentrations shift to reach equilibrium. First, calculate the reaction quotient

Q_c

using the initial concentrations:

Q_c = \frac{[N_2O_4]}{[NO_2]^2} = \frac{0.010}{(0.040)^2} = 6.25

. Since

Q_c = 6.25 < K_c = 170

, the reaction must shift right (toward products) to reach equilibrium. Set up an ICE table. Let

x

be the amount of

NO_2

that reacts:

$[NO_2]$ changes from 0.040 to $(0.040 - 2x)$
$[N_2O_4]$ changes from 0.010 to $(0.010 + x)$

At equilibrium:

K_c = \frac{0.010 + x}{(0.040 - 2x)^2} = 170

Expanding:

0.010 + x = 170(0.040 - 2x)^2 = 170(0.0016 - 0.16x + 4x^2)

This gives:

680x^2 - 27.2x + 0.262 = 0

Using the quadratic formula:

x = 0.012

x = 0.0321

Since

x = 0.0321

would make

[NO_2]

negative,

x = 0.012

. Therefore:

[NO_2] = 0.040 - 2(0.012) = 0.016

M Answer B (0.016 M) is correct. Answer A (0.012 M) represents the value of

x

, not the final concentration. Answer C (0.020 M) is the initial concentration. Answer D (0.024 M) incorrectly adds

2x

instead of subtracting it. Remember: always check that

Q_c

versus

K_c

to determine reaction direction before setting up your ICE table, and verify your final answer makes chemical sense.

Question 5

At equilibrium for $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ , the concentrations are [ $SO_2$ ] = 0.60 M, [ $O_2$ ] = 0.30 M, and [ $SO_3$ ] = 1.2 M. If the volume is suddenly halved while temperature remains constant, what will be the new equilibrium concentration of $SO_3$ ?

2.1 M
2.3 M
2.5 M
2.7 M (correct answer)
2.9 M

Explanation: When you encounter equilibrium problems involving volume changes, you need to consider how the change affects the reaction quotient and which direction the equilibrium will shift to minimize the disturbance. First, calculate the initial equilibrium constant:

K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]} = \frac{(1.2)^2}{(0.60)^2(0.30)} = \frac{1.44}{0.108} = 13.3

When volume is halved, all concentrations double immediately: [

SO_2

] = 1.20 M, [

O_2

] = 0.60 M, [

SO_3

] = 2.4 M. Now calculate the reaction quotient:

Q_c = \frac{(2.4)^2}{(1.20)^2(0.60)} = \frac{5.76}{0.864} = 6.67

Since

Q_c < K_c

, the reaction shifts right to re-establish equilibrium. Set up an ICE table with the new concentrations and let x = moles/L of

SO_2

that react. At equilibrium:

13.3 = \frac{(2.4 + 2x)^2}{(1.20 - 2x)^2(0.60 - x)}

Solving this equation (which requires iterative methods or approximation) gives x ≈ 0.15, making [

SO_3

] = 2.4 + 2(0.15) = 2.7 M. Choice A (2.1 M) incorrectly assumes the reaction shifts left. Choice B (2.3 M) likely comes from insufficient equilibrium shift calculation. Choice C (2.5 M) represents an intermediate value that doesn't satisfy the equilibrium expression. Study tip: For volume changes in equilibrium problems, always remember that decreasing volume favors the side with fewer gas molecules. Here, the left side has 3 moles of gas versus 2 on the right, so compression drives the reaction toward products.

Question 6

For the equilibrium $2IBr(g) \rightleftharpoons I_2(g) + Br_2(g)$ , $K_c = 8.5 \times 10^{-3}$ at 150°C. A sealed container initially holds 0.500 M $IBr$ . After equilibrium is established, what fraction of the original $IBr$ remains unreacted?

0.75
0.80
0.85 (correct answer)
0.90
0.95

Explanation: This question tests equilibrium calculations using an ICE table (Initial, Change, Equilibrium) to determine how much reactant remains after reaching equilibrium. Start by setting up your ICE table. Initially, you have 0.500 M

IBr

and 0 M of both products. Let

x

represent the moles per liter of

IBr

that react. From the stoichiometry,

2x

moles of

IBr

are consumed to produce

x

moles each of

I_2

and

Br_2

. At equilibrium:

[IBr] = 0.500 - 2x

[I_2] = x

[Br_2] = x

Substitute into the equilibrium expression:

K_c = \frac{[I_2][Br_2]}{[IBr]^2} = \frac{x \cdot x}{(0.500 - 2x)^2} = 8.5 \times 10^{-3}

This gives you:

\frac{x^2}{(0.500 - 2x)^2} = 8.5 \times 10^{-3}

Taking the square root:

\frac{x}{0.500 - 2x} = 0.0922

Solving:

x = 0.0922(0.500 - 2x) = 0.0461 - 0.1844x

1.1844x = 0.0461

, so

x = 0.0389

The equilibrium concentration of

IBr

0.500 - 2(0.0389) = 0.422

M. The fraction remaining is

\frac{0.422}{0.500} = 0.844 ≈ 0.85

, which is answer C. Answer A (0.75) assumes too much reaction occurred. Answer B (0.80) and D (0.90) represent calculation errors, likely from incorrect stoichiometry or algebraic mistakes. Remember: always double-check your equilibrium expression matches the balanced equation's stoichiometry, and verify your final answer makes chemical sense given the

K_c

value.

Question 7

The reaction $CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)$ has $K_c = 4.0$ at 1000 K. A mixture initially contains 0.300 M $CO$ , 0.300 M $H_2O$ , 0.200 M $CO_2$ , and 0.100 M $H_2$ . What is the equilibrium concentration of $CO_2$ ?

0.235 M
0.255 M
0.275 M (correct answer)
0.295 M
0.315 M

Explanation: When you encounter an equilibrium problem with given initial concentrations and a known

K_c

, you need to determine whether the system is at equilibrium and, if not, predict which direction it will shift. First, calculate the reaction quotient

Q_c

using the initial concentrations:

Q_c = \frac{[CO_2][H_2]}{[CO][H_2O]} = \frac{(0.200)(0.100)}{(0.300)(0.300)} = \frac{0.0200}{0.0900} = 0.222

Since

Q_c = 0.222 < K_c = 4.0

, the reaction must shift right (toward products) to reach equilibrium. Set up an ICE table with

x

representing the molar change. At equilibrium:

$[CO] = 0.300 - x$
$[H_2O] = 0.300 - x$
$[CO_2] = 0.200 + x$
$[H_2] = 0.100 + x$

Substitute into the equilibrium expression:

4.0 = \frac{(0.200 + x)(0.100 + x)}{(0.300 - x)(0.300 - x)}

Expanding and solving this quadratic equation yields

x = 0.075

M. Therefore:

[CO_2] = 0.200 + 0.075 = 0.275

M, which is answer C. Answer A (0.235 M) corresponds to

x = 0.035

, Answer B (0.255 M) to

x = 0.055

, and Answer D (0.295 M) to

x = 0.095

. These represent calculation errors in solving the quadratic equation or mistakes in the algebraic manipulation. Remember: always check whether

Q_c

equals

K_c

first—this tells you the reaction direction and confirms your final answer makes sense.

Question 8

The equilibrium $A(g) + 2B(g) \rightleftharpoons C(g) + D(g)$ has $K_c = 25$ at 400 K. If the equilibrium concentrations are [A] = 0.10 M, [B] = 0.20 M, and [C] = 0.50 M, what is the equilibrium concentration of D?

0.15 M
0.20 M (correct answer)
0.25 M
0.30 M
0.35 M

Explanation: When you encounter an equilibrium problem with a given

K_c

value and partial concentration data, you're working with the equilibrium constant expression. For any reaction,

K_c

equals the product of product concentrations (raised to their stoichiometric coefficients) divided by the product of reactant concentrations (also raised to their coefficients). For this reaction, the equilibrium expression is:

K_c = \frac{[C][D]}{[A][B]^2}

Since

K_c = 25

and you know [A] = 0.10 M, [B] = 0.20 M, and [C] = 0.50 M, you can solve for [D]:

25 = \frac{(0.50)[D]}{(0.10)(0.20)^2}

25 = \frac{(0.50)[D]}{(0.10)(0.04)}

25 = \frac{(0.50)[D]}{0.004}

[D] = \frac{25 \times 0.004}{0.50} = \frac{0.10}{0.50} = 0.20 \text{ M}

This confirms answer B is correct. Answer A (0.15 M) likely comes from calculation errors in the arithmetic. Answer C (0.25 M) might result from incorrectly using

[B]^1

instead of

[B]^2

in the denominator, ignoring the stoichiometric coefficient. Answer D (0.30 M) could arise from setting up the equilibrium expression incorrectly or making multiple arithmetic mistakes. Always write the

K_c

expression first, then substitute known values systematically. Double-check that you're using the correct stoichiometric coefficients as exponents—this is where many students make errors in equilibrium calculations.

Question 9

Consider the equilibrium $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ with $K_p = 6.0 \times 10^{-2}$ at 500°C. If the equilibrium partial pressures are $P_{N_2} = 3.0$ atm and $P_{H_2} = 1.0$ atm, what is the partial pressure of $NH_3$ ?

0.38 atm
0.42 atm (correct answer)
0.46 atm
0.50 atm
0.54 atm

Explanation: When you encounter equilibrium problems with partial pressures, you're working with the equilibrium constant

K_p

, which relates the partial pressures of products and reactants at equilibrium. For this reaction,

K_p = \frac{P_{NH_3}^2}{P_{N_2} \times P_{H_2}^3}

. To find the partial pressure of

NH_3

, substitute the known values into the

K_p

expression:

6.0 \times 10^{-2} = \frac{P_{NH_3}^2}{(3.0) \times (1.0)^3}

. This simplifies to

6.0 \times 10^{-2} = \frac{P_{NH_3}^2}{3.0}

. Solving for

P_{NH_3}^2

P_{NH_3}^2 = 6.0 \times 10^{-2} \times 3.0 = 0.18

. Taking the square root gives

P_{NH_3} = 0.42

atm. Choice A (0.38 atm) likely results from calculation errors, possibly forgetting to multiply by 3.0 or making arithmetic mistakes. Choice C (0.46 atm) might come from incorrectly setting up the

K_p

expression or rounding errors during calculation. Choice D (0.50 atm) could result from misplacing exponents in the equilibrium expression or using an incorrect mathematical approach. Remember that

K_p

expressions always follow the pattern: products over reactants, each raised to their stoichiometric coefficients. Double-check your exponents match the balanced equation, and be careful with arithmetic when dealing with scientific notation. Practice setting up these expressions systematically to avoid mistakes.

Question 10

The equilibrium $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$ has $K_p = 0.0211$ at 250°C. If the initial pressure of $PCl_5$ is 0.500 atm in an evacuated container, what is the equilibrium pressure of $PCl_3$ ?

0.095 atm
0.115 atm (correct answer)
0.135 atm
0.155 atm
0.175 atm

Explanation: When you encounter equilibrium problems involving gas-phase reactions, you need to set up an ICE table (Initial, Change, Equilibrium) and use the equilibrium constant expression to solve for unknown pressures. For this reaction, start by setting up your ICE table. Initially, you have 0.500 atm of

PCl_5

and 0 atm of both products. Let x represent the amount of

PCl_5

that dissociates. At equilibrium:

PCl_5

= (0.500 - x) atm,

PCl_3

= x atm, and

Cl_2

= x atm. The equilibrium expression is:

K_p = \frac{P_{PCl_3} \times P_{Cl_2}}{P_{PCl_5}} = \frac{x \times x}{0.500 - x} = \frac{x^2}{0.500 - x}

Substituting

K_p = 0.0211

0.0211 = \frac{x^2}{0.500 - x}

Rearranging:

0.0211(0.500 - x) = x^2

, which gives

x^2 + 0.0211x - 0.01055 = 0

Using the quadratic formula:

x = \frac{-0.0211 + \sqrt{(0.0211)^2 + 4(0.01055)}}{2} = 0.115

atm This makes (B) 0.115 atm correct. Choice (A) 0.095 atm likely results from calculation errors in the quadratic formula. Choice (C) 0.135 atm might come from neglecting the -x term in the denominator, treating it as simply 0.500. Choice (D) 0.155 atm could result from sign errors or incorrect setup of the equilibrium expression. Remember: always double-check your quadratic equation setup and verify your answer makes physical sense—the equilibrium pressure can't exceed the initial pressure of reactant.

Question 11

For the reaction $2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g)$ , $K_c = 1.6 \times 10^{-5}$ at 35°C. If 0.500 mol $NOCl$ is placed in a 2.0 L container, what is the equilibrium concentration of $NO$ ?

0.0085 M
0.0105 M
0.0125 M (correct answer)
0.0145 M
0.0165 M

Explanation: When you encounter an equilibrium problem with a given equilibrium constant, you need to set up an ICE table (Initial, Change, Equilibrium) and solve using the equilibrium expression. Start with the initial concentration of NOCl:

\frac{0.500 \text{ mol}}{2.0 \text{ L}} = 0.250 \text{ M}

. Set up your ICE table with NOCl starting at 0.250 M and products at 0. Let x represent the change in concentration. At equilibrium: [NOCl] = 0.250 - 2x, [NO] = 2x, [Cl₂] = x The equilibrium expression is:

K_c = \frac{[NO]^2[Cl_2]}{[NOCl]^2} = 1.6 \times 10^{-5}

Substituting:

\frac{(2x)^2(x)}{(0.250-2x)^2} = 1.6 \times 10^{-5}

Since Kc is very small, assume 2x << 0.250, so the denominator becomes approximately (0.250)². This gives:

\frac{4x^3}{(0.250)^2} = 1.6 \times 10^{-5}

Solving:

x^3 = \frac{1.6 \times 10^{-5} \times 0.0625}{4} = 2.5 \times 10^{-7}

Therefore:

x = 6.3 \times 10^{-3}

and [NO] = 2x = 0.0126 M ≈ 0.0125 M Answer C (0.0125 M) is correct. Answer A (0.0085 M) likely results from calculation errors in the cubic root. Answer B (0.0105 M) and D (0.0145 M) represent common arithmetic mistakes or incorrect assumption handling. Remember: for small equilibrium constants, use the approximation method to simplify calculations, but always verify your assumption holds true at the end.