Kinetics and Equilibrium | AP - Wyatt's Notes

Chemical Kinetics (CED Unit 5)

Chemical kinetics studies the rates of chemical reactions and the factors that affect them.

Rate of Reaction

The average rate of a reaction is the change in concentration of a reactant or product per unit Time:

\mathrm{Rate = -\frac{1}{a}\frac{\Delta[\mathrm{A]}{\Delta t} = \frac{1}{b}\frac{\Delta[\mathrm{B]}{\Delta t}

Where $a$ and $b$ are stoichiometric coefficients.

Rate Law

For the reaction $a\mathrm{A + b\mathrm{B \to \mathrm{products$ :

\mathrm{Rate = k[\mathrm{A]^m[\mathrm{B]^n

$k$ is the rate constant (depends on temperature).
$m$ is the order with respect to A (determined experimentally, NOT from the stoichiometric coefficient).
$n$ is the order with respect to B.
The overall order is $m + n$ .

Determining the Rate Law from Initial Rate Data

Compare experiments where only one concentration changes. If doubling [A] doubles the rate, the Reaction is first order in A. If doubling [A] quadruples the rate, it is second order in A.

Example Determine the rate law from the following data for $2\mathrm{A + \mathrm{B \to \mathrm{C$ :

Experiment	[A] (M)	[B] (M)	Initial Rate (M/s)
1	0.10	0.10	0.0020
2	0.20	0.10	0.0040
3	0.10	0.20	0.0080

Comparing 1 and 2: [A] doubles, rate doubles $\implies$ first order in A ( $m = 1$ ).

Comparing 1 and 3: [B] doubles, rate quadruples $\implies$ second order in B ( $n = 2$ ).

\mathrm{Rate = k[\mathrm{A][\mathrm{B]^2

Find $k$ from experiment 1: $0.0020 = k(0.10)(0.10)^2 \implies k = 2.0 \mathrm{ M^{-2}\mathrm{s^{-1}$ .

Worked Example: Rate Law Determination

Determine the rate law from the following data for $\mathrm{A + \mathrm{B \to \mathrm{C$ :

Experiment	[A] (M)	[B] (M)	Rate (M/s)
1	0.10	0.20	0.0040
2	0.20	0.20	0.0080
3	0.20	0.40	0.0080
4	0.40	0.20	0.0160

Comparing 1 and 2: [A] doubles, rate doubles. First order in A.

Comparing 2 and 3: [B] doubles, rate unchanged. Zero order in B.

Comparing 2 and 4: [A] doubles, rate doubles. Confirms first order in A.

$\mathrm{Rate = k[\mathrm{A], \quad k = \frac{0.0080}{0.20} = 0.040 \mathrm{ s^{-1}$

Integrated Rate Laws (CED Unit 5)

First-Order Reactions

\ln[\mathrm{A]_t = -kt + \ln[\mathrm{A]_0

Or equivalently:

[\mathrm{A]_t = [\mathrm{A]_0 e^{-kt}

Half-life: $\displaystyle t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$

The half-life is independent of initial concentration.

Second-Order Reactions

\frac{1}{[\mathrm{A]_t} = kt + \frac{1}{[\mathrm{A]_0}

Half-life: $\displaystyle t_{1/2} = \frac{1}{k[\mathrm{A]_0}$

The half-life depends on initial concentration.

Zero-Order Reactions

[\mathrm{A]_t = -kt + [\mathrm{A]_0

Half-life: $\displaystyle t_{1/2} = \frac{[\mathrm{A]_0}{2k}$

Identifying the Order

Order	Linear Plot	Slope
Zero	[A] vs $t$	$-k$
First	$\ln[\mathrm{A]$ vs $t$	$-k$
Second	$1/[\mathrm{A]$ vs $t$	$k$

Example The decomposition of $\mathrm{N_2\mathrm{O_5$ is first order with $k = 5.0 \times 10^{-4} \mathrm{ s^{-1}$ . How long does it take for 80% of a $0.500 \mathrm{ M$ Sample to decompose?

$[\mathrm{A]_t = 0.20 \times 0.500 = 0.100 \mathrm{ M$ .

\ln(0.100) = -(5.0 \times 10^{-4})t + \ln(0.500)

-2.303 = -5.0 \times 10^{-4} t + (-0.693)

-2.303 + 0.693 = -5.0 \times 10^{-4} t

T = \frac{-1.610}{-5.0 \times 10^{-4}} = 3220 \mathrm{ s \approx 53.7 \mathrm{ min

Worked Example: Half-Life Calculation

A first-order reaction has a rate constant of $0.030 \mathrm{ min^{-1}$ . Calculate the half-life And the time for 75% decomposition.

$t_{1/2} = \frac{0.693}{0.030} = 23.1 \mathrm{ min$

After 75% decomposition, $[\mathrm{A]_t = 0.25[\mathrm{A]_0$ :

$t = \frac{1}{k}\ln\frac{[\mathrm{A]_0}{[\mathrm{A]_t} = \frac{1}{0.030}\ln 4 = 33.3 \times 1.386 = 46.2 \mathrm{ min$

Note that 75% decomposition takes two half-lives (46.2 min $\approx 2 \times 23.1$ min).

Reaction Mechanisms (CED Unit 5)

A reaction mechanism is a sequence of elementary steps that sum to the overall reaction.

Molecularity

Molecularity	Description	Rate Law
Unimolecular	One molecule reacts	Rate = $k[\mathrm{A]$
Bimolecular	Two molecules collide	Rate = $k[\mathrm{A][\mathrm{B]$
Termolecular	Three molecules collide	Rate = $k[\mathrm{A][\mathrm{B][\mathrm{C]$ (rare)

Rate-Determining Step

The slowest step in the mechanism determines the overall rate law.

Requirements for a Valid Mechanism

The elementary steps must sum to the overall reaction.
The rate law derived from the rate-determining step must agree with the experimental rate law.
If a fast equilibrium precedes the slow step, use the equilibrium approximation to express intermediates in terms of reactants.

Example The reaction $2\mathrm{NO_2 + \mathrm{F_2 \to 2\mathrm{NO_2\mathrm{F$ has the experimental rate Law $\mathrm{Rate = k[\mathrm{NO_2][\mathrm{F_2]$ .

Proposed mechanism:

Step 1 (slow): $\mathrm{NO_2 + \mathrm{F_2 \to \mathrm{NO_2\mathrm{F + \mathrm{F$

Step 2 (fast): $\mathrm{F + \mathrm{NO_2 \to \mathrm{NO_2\mathrm{F$

The rate law from the slow step: $\mathrm{Rate = k[\mathrm{NO_2][\mathrm{F_2]$ . This matches the Experimental rate law, so the mechanism is plausible.

Sum: $2\mathrm{NO_2 + \mathrm{F_2 \to 2\mathrm{NO_2\mathrm{F$ .

Worked Example: Mechanism with Fast Equilibrium

The reaction $2\mathrm{NO + \mathrm{O_2 \to 2\mathrm{NO_2$ has the experimental rate law $\mathrm{Rate = k[\mathrm{NO]^2[\mathrm{O_2]$ .

Proposed mechanism:

Step 1 (fast equilibrium): $\mathrm{NO + \mathrm{NO \rightleftharpoons \mathrm{N_2\mathrm{O_2$

Step 2 (slow): $\mathrm{N_2\mathrm{O_2 + \mathrm{O_2 \to 2\mathrm{NO_2$

From step 1: $K = \frac{[\mathrm{N_2\mathrm{O_2]}{[\mathrm{NO]^2}$ So $[\mathrm{N_2\mathrm{O_2] = K[\mathrm{NO]^2$ .

Rate from step 2: $\mathrm{Rate = k_2[\mathrm{N_2\mathrm{O_2][\mathrm{O_2] = k_2 K[\mathrm{NO]^2[\mathrm{O_2]$ .

This matches the experimental rate law with $k = k_2 K$ .

Arrhenius Equation (CED Unit 5)

The rate constant depends on temperature:

K = A e^{-E_a/(RT)}

In logarithmic form:

\ln k = -\frac{E_a}{R}\cdot\frac{1}{T} + \ln A

Where $E_a$ is the activation energy, $A$ is the frequency factor, and $R = 8.314 \mathrm{ J/(mol\cdot\mathrm{K)$ .

Two-Point Form

\ln\frac{k_2}{k_1} = \frac{E_a}{R}\!\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

Catalysis

A catalyst provides an alternative pathway with a lower activation energy. It increases the rate Constant $k$ but is not consumed in the reaction.

Homogeneous catalyst: same phase as the reactants.
Heterogeneous catalyst: different phase ( a solid surface).

Example A reaction has $k = 3.0 \times 10^{-3} \mathrm{ s^{-1}$ at $300 \mathrm{ K$ and $k = 2.4 \times 10^{-2} \mathrm{ s^{-1}$ at $350 \mathrm{ K$ . Find $E_a$ .

\ln\frac{2.4 \times 10^{-2}}{3.0 \times 10^{-3}} = \frac{E_a}{8.314}\!\left(\frac{1}{300} - \frac{1}{350}\right)

\ln 8.0 = \frac{E_a}{8.314}\!\left(\frac{50}{105000}\right)

2.079 = \frac{E_a}{8.314}(4.762 \times 10^{-4})

E_a = \frac{2.079}{4.762 \times 10^{-4}} \times 8.314 = 36300 \mathrm{ J/mol = 36.3 \mathrm{ kJ/mol

Worked Example: Effect of a Catalyst

A reaction has $E_a = 75 \mathrm{ kJ/mol$ without a catalyst and $E_a = 50 \mathrm{ kJ/mol$ with a Catalyst. Calculate the ratio of rate constants at $298 \mathrm{ K$ .

$\frac{k_{\mathrm{cat}}{k_{\mathrm{uncat}} = \frac{A e^{-50000/(8.314 \times 298)}}{A e^{-75000/(8.314 \times 298)}} = e^{(75000 - 50000)/(8.314 \times 298)}$

$= e^{25000/2478} = e^{10.09} = 2.4 \times 10^4$

The catalyst increases the rate constant by a factor of about 24,000 at room temperature.

Chemical Equilibrium (CED Unit 7)

Dynamic Equilibrium

At equilibrium, the forward and reverse reaction rates are equal. Concentrations of reactants and Products remain constant (but not necessarily equal).

The Equilibrium Constant ( $K$ )

For the reaction $a\mathrm{A + b\mathrm{B \rightleftharpoons c\mathrm{C + d\mathrm{D$ :

K_c = \frac{[\mathrm{C]^c[\mathrm{D]^d}{[\mathrm{A]^a[\mathrm{B]^b}

For gas-phase reactions, $K_p$ uses partial pressures:

K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}

Relationship Between $K_c$ and $K_p$

K_p = K_c(RT)^{\Delta n}

Where $\Delta n = (\mathrm{moles gaseous products) - (\mathrm{moles gaseous reactants)$ .

Heterogeneous Equilibria

Pure solids and pure liquids are not included in the equilibrium expression because their activities Are constant (equal to 1).

Example Write the equilibrium expression for:

\mathrm{CaCO_3(s) \rightleftharpoons \mathrm{CaO(s) + \mathrm{CO_2(g)

K_p = P_{\mathrm{CO_2}

Only $\mathrm{CO_2$ appears because solids are omitted.

The Reaction Quotient ( $Q$ )

Q = \frac{[\mathrm{C]^c[\mathrm{D]^d}{[\mathrm{A]^a[\mathrm{B]^b}

$Q$ has the same form as $K$ but uses current (non-equilibrium) concentrations.

Comparison	Result
$Q \lt K$	Reaction proceeds forward
$Q = K$	System is at equilibrium
$Q \gt K$	Reaction proceeds in reverse

Le Chatelier”s Principle (CED Unit 7)

If a stress is applied to a system at equilibrium, the system shifts to counteract the stress.

Types of Stress

Stress	Shift Direction	Effect on $K$
Add reactant	Toward products	None
Add product	Toward reactants	None
Remove reactant	Toward reactants	None
Remove product	Toward products	None
Increase temperature (endothermic)	Toward products	Increases
Increase temperature (exothermic)	Toward reactants	Decreases
Increase pressure (by decreasing volume)	Toward fewer moles of gas	None
Add catalyst	No shift	None

Key point: Only temperature changes affect the value of $K$ .

Example For $\mathrm{H_2(g) + \mathrm{I_2(g) \rightleftharpoons 2\mathrm{HI(g)$ with $K_c = 50.5$ at $448^\circ\mathrm{C$ :

If $1.00 \mathrm{ M \mathrm{H_2$ and $1.00 \mathrm{ M \mathrm{I_2$ are mixed, find [HI] at Equilibrium.

Species	Initial	Change	Equilibrium
$\mathrm{H_2$	1.00	$-x$	$1.00 - x$
$\mathrm{I_2$	1.00	$-x$	$1.00 - x$
$\mathrm{HI$	0	$+2x$	$2x$

K_c = \frac{(2x)^2}{(1.00 - x)^2} = 50.5

\frac{2x}{1.00 - x} = \sqrt{50.5} = 7.11

2x = 7.11(1.00 - x) \implies 2x = 7.11 - 7.11x \implies 9.11x = 7.11 \implies x = 0.781

[\mathrm{HI] = 2(0.781) = 1.56 \mathrm{ M

Worked Example: ICE Table with Quadratic

For $\mathrm{PCl_5(g) \rightleftharpoons \mathrm{PCl_3(g) + \mathrm{Cl_2(g)$ , $K_c = 0.0420$ at $500 \mathrm{ K$ . If $2.00 \mathrm{ M \mathrm{PCl_5$ is placed in a flask, find all equilibrium Concentrations.

Species	Initial	Change	Equilibrium
$\mathrm{PCl_5$	2.00	$-x$	$2.00 - x$
$\mathrm{PCl_3$	0	$+x$	$x$
$\mathrm{Cl_2$	0	$+x$	$x$

$K_c = \frac{x^2}{2.00 - x} = 0.0420$

$x^2 = 0.0420(2.00 - x) = 0.0840 - 0.0420x$

$x^2 + 0.0420x - 0.0840 = 0$

Using the quadratic formula: $x = \frac{-0.0420 + \sqrt{0.0420^2 + 4(0.0840)}}{2} = \frac{-0.0420 + 0.581}{2} = 0.269 \mathrm{ M$

$[\mathrm{PCl_5] = 1.73 \mathrm{ M$ , $[\mathrm{PCl_3] = 0.269 \mathrm{ M$ $[\mathrm{Cl_2] = 0.269 \mathrm{ M$ .

Worked Example: $K_c$ to $K_p$ Conversion

For $\mathrm{N_2(g) + 3\mathrm{H_2(g) \rightleftharpoons 2\mathrm{NH_3(g)$ at $400^{\circ}\mathrm{C$ , $K_c = 0.500$ . Calculate $K_p$ .

$\Delta n = 2 - (1 + 3) = -2$

$K_p = K_c(RT)^{\Delta n} = 0.500 \times (0.08206 \times 673)^{-2} = \frac{0.500}{(55.23)^2} = \frac{0.500}{3049} = 1.64 \times 10^{-4}$

The small $K_p$ reflects the fact that the equilibrium lies to the left at this temperature (the Haber process is run at higher temperatures to increase $K$ ).

Solubility Equilibrium (CED Unit 7)

Solubility Product Constant ( $K_{sp}$ )

For a sparingly soluble salt $\mathrm{M_a\mathrm{X_b(s) \rightleftharpoons a\mathrm{M^{b+}(aq) + b\mathrm{X^{a-}(aq)$ :

K_{sp} = [\mathrm{M^{b+}]^a[\mathrm{X^{a-}]^b

Common Ion Effect

The solubility of a salt decreases when a common ion is already present in solution.

Predicting Precipitation

Compare $Q_{sp}$ with $K_{sp}$ :

$Q_{sp} \lt K_{sp}$ : no precipitate (unsaturated)
$Q_{sp} = K_{sp}$ : saturated, at equilibrium
$Q_{sp} \gt K_{sp}$ : precipitate forms

Example Will a precipitate form when $50.0 \mathrm{ mL$ of $0.0010 \mathrm{ M \mathrm{Pb(NO_3)_2$ is Mixed with $50.0 \mathrm{ mL$ of $0.0020 \mathrm{ M \mathrm{NaCl$ ? $K_{sp}(\mathrm{PbCl_2) = 1.7 \times 10^{-5}$ .

After mixing (volumes double):

$[\mathrm{Pb^{2+}] = 0.00050 \mathrm{ M$ , $[\mathrm{Cl^-] = 0.0010 \mathrm{ M$ .

Q_{sp} = (0.00050)(0.0010)^2 = 5.0 \times 10^{-10}

Since $Q_{sp} = 5.0 \times 10^{-10} \lt K_{sp} = 1.7 \times 10^{-5}$ No precipitate forms.

Worked Example: $K_{sp}$ Calculation from Solubility

The solubility of $\mathrm{AgCl$ in water at $25^{\circ}\mathrm{C$ is $1.3 \times 10^{-5} \mathrm{ M$ . Calculate $K_{sp}$ .

$\mathrm{AgCl(s) \rightleftharpoons \mathrm{Ag^+(aq) + \mathrm{Cl^-(aq)$

$K_{sp} = [\mathrm{Ag^+][\mathrm{Cl^-] = (1.3 \times 10^{-5})^2 = 1.7 \times 10^{-10}$

Worked Example: Common Ion Effect

Calculate the solubility of $\mathrm{AgCl$ in $0.10 \mathrm{ M \mathrm{NaCl$ . $K_{sp} = 1.7 \times 10^{-10}$ .

$K_{sp} = s(s + 0.10) \approx s \times 0.10$

$s = \frac{1.7 \times 10^{-10}}{0.10} = 1.7 \times 10^{-9} \mathrm{ M$

Compare with solubility in pure water: $s_0 = \sqrt{1.7 \times 10^{-10}} = 1.3 \times 10^{-5} \mathrm{ M$ .

The common ion effect reduces solubility by a factor of about 7,600.

flowchart TD
    A[5_Kinetics And Equilibrium] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary Table: Rate Laws

Order	Integrated Law	Half-Life	Units of $k$	Linear Plot
0	$[\mathrm{A] = -kt + [\mathrm{A]_0$	$[\mathrm{A]_0/(2k)$	$\mathrm{M s^{-1}$	$[\mathrm{A]$ vs $t$
1	$\ln[\mathrm{A] = -kt + \ln[\mathrm{A]_0$	$0.693/k$	$\mathrm{s^{-1}$	$\ln[\mathrm{A]$ vs $t$
2	$1/[\mathrm{A] = kt + 1/[\mathrm{A]_0$	$1/(k[\mathrm{A]_0)$	$\mathrm{M^{-1}\mathrm{s^{-1}$	$1/[\mathrm{A]$ vs $t$

Worked Example: Mechanism with Fast Equilibrium

The reaction $2\mathrm{NO + \mathrm{O_2 \to 2\mathrm{NO_2$ has the experimental rate law $\mathrm{Rate = k[\mathrm{NO]^2[\mathrm{O_2]$ .

Proposed mechanism:

Step 1 (fast equilibrium): $\mathrm{NO + \mathrm{NO \rightleftharpoons \mathrm{N_2\mathrm{O_2$

Step 2 (slow): $\mathrm{N_2\mathrm{O_2 + \mathrm{O_2 \to 2\mathrm{NO_2$

From step 1: $K = \frac{[\mathrm{N_2\mathrm{O_2]}{[\mathrm{NO]^2}$ So $[\mathrm{N_2\mathrm{O_2] = K[\mathrm{NO]^2$ .

Rate from step 2: $\mathrm{Rate = k_2[\mathrm{N_2\mathrm{O_2][\mathrm{O_2] = k_2 K[\mathrm{NO]^2[\mathrm{O_2]$ .

This matches the experimental rate law with $k = k_2 K$ .

Summary Table: Rate Laws

Order	Integrated Law	Half-Life	Units of $k$	Linear Plot
0	$[\mathrm{A] = -kt + [\mathrm{A]_0$	$[\mathrm{A]_0/(2k)$	$\mathrm{M s^{-1}$	$[\mathrm{A]$ vs $t$
1	$\ln[\mathrm{A] = -kt + \ln[\mathrm{A]_0$	$0.693/k$	$\mathrm{s^{-1}$	$\ln[\mathrm{A]$ vs $t$
2	$1/[\mathrm{A] = kt + 1/[\mathrm{A]_0$	$1/(k[\mathrm{A]_0)$	$\mathrm{M^{-1}\mathrm{s^{-1}$	$1/[\mathrm{A]$ vs $t$

Summary Table: Factors Affecting Reaction Rate

Factor	Effect on Rate	Explanation
Concentration	Increases	More collisions per unit time
Temperature	Increases	More molecules have energy $\geq E_a$ ; $k$ increases exponentially
Surface area	Increases	More exposed particles for collision
Catalyst	Increases	Lowers $E_a$ by providing an alternative pathway
Pressure (gas)	Increases	Higher concentration = more collisions
Light (photo)	May increase	Provides energy to overcome $E_a$ via photon absorption

Summary Table: Le Chatelier’s Principle

Stress Applied	Direction of Shift	Effect on $K$
Add reactant	Toward products	None
Add product	Toward reactants	None
Remove reactant	Toward reactants	None
Remove product	Toward products	None
Increase temperature (endo.)	Toward products	Increases
Increase temperature (exo.)	Toward reactants	Decreases
Decrease volume	Fewer gas moles	None
Add catalyst	No shift	None

Common Pitfalls

Confusing rate law orders with stoichiometric coefficients. The orders must be determined experimentally.
Using the wrong integrated rate law. Identify the order first from the data or from the problem statement.
Forgetting that catalysts do not change $K$ . Catalysts increase the rate of both forward and reverse reactions equally.
Including solids and liquids in $K$ expressions. Only gases and aqueous species appear.
Incorrectly predicting the effect of pressure changes. Only changes in the number of gas moles matter. Adding an inert gas at constant volume does not shift equilibrium.
Making algebraic errors in ICE tables. Check that the stoichiometric coefficients match the changes.
Forgetting to account for dilution when mixing solutions (total volume changes).
Confusing $Q$ with $K$ . $Q$ uses current concentrations; $K$ uses equilibrium concentrations.
Using the quadratic formula incorrectly in ICE table problems. Always take the positive root for $x$ (concentrations cannot be negative).
Forgetting that $K_p$ and $K_c$ are related by $(RT)^{\Delta n}$ . Use the correct $\Delta n$ .

Practice Questions

For a first-order reaction with $k = 0.050 \mathrm{ s^{-1}$ How long does it take for the concentration to decrease from $0.80 \mathrm{ M$ to $0.20 \mathrm{ M$ ?
The following data were collected for the reaction $\mathrm{A + \mathrm{B \to \mathrm{C$ :

[A] (M)	[B] (M)	Rate (M/s)
0.10	0.10	0.0030
0.20	0.10	0.0060
0.20	0.20	0.0240

Determine the rate law and rate constant.

At $400 \mathrm{ K$ , $k = 6.4 \times 10^{-3} \mathrm{ M^{-1}\mathrm{s^{-1}$ . At $450 \mathrm{ K$ , $k = 3.2 \times 10^{-2} \mathrm{ M^{-1}\mathrm{s^{-1}$ . Find $E_a$ .
For $\mathrm{PCl_5(g) \rightleftharpoons \mathrm{PCl_3(g) + \mathrm{Cl_2(g)$ , $K_p = 1.80$ at $250^\circ\mathrm{C$ . If $0.500 \mathrm{ atm$ of $\mathrm{PCl_5$ is placed in a flask, find the equilibrium partial pressures of all species.
Does a precipitate form when $100 \mathrm{ mL$ of $0.010 \mathrm{ M \mathrm{AgNO_3$ is mixed with $100 \mathrm{ mL$ of $0.010 \mathrm{ M \mathrm{NaCl$ ? $K_{sp}(\mathrm{AgCl) = 1.8 \times 10^{-10}$ .
Explain how Le Chatelier’s principle applies when the volume of the container is decreased for the reaction $\mathrm{N_2(g) + 3\mathrm{H_2(g) \rightleftharpoons 2\mathrm{NH_3(g)$ .
For a reaction with $\Delta H = -92 \mathrm{ kJ/mol$ What happens to $K$ when the temperature increases from $298 \mathrm{ K$ to $400 \mathrm{ K$ ?
Calculate the molar solubility of $\mathrm{PbSO_4$ in pure water and in $0.10 \mathrm{ M \mathrm{Na_2\mathrm{SO_4$ . $K_{sp}(\mathrm{PbSO_4) = 1.6 \times 10^{-8}$ .
The half-life of a reaction is $120 \mathrm{ s$ and the initial concentration is $0.50 \mathrm{ M$ . If the reaction is first order, what is the rate constant? What is the concentration after $240 \mathrm{ s$ ?
Write the equilibrium expression for $\mathrm{BaSO_4(s) \rightleftharpoons \mathrm{Ba^{2+}(aq) + \mathrm{SO_4^{2-}(aq)$ and calculate the concentration of $\mathrm{Ba^{2+}$ in a saturated solution. $K_{sp} = 1.1 \times 10^{-10}$ .
A proposed mechanism for a reaction is: Step 1 (fast): $\mathrm{NO(g) + \mathrm{Br_2(g) \rightleftharpoons \mathrm{NOBr_2(g)$ Step 2 (slow): $\mathrm{NOBr_2(g) + \mathrm{NO(g) \to 2\mathrm{NOBr(g)$ Derive the rate law from this mechanism.
Calculate the solubility of $\mathrm{PbI_2$ in $0.020 \mathrm{ M \mathrm{KI$ . $K_{sp}(\mathrm{PbI_2) = 7.9 \times 10^{-9}$ .
For the reaction $2\mathrm{SO_2(g) + \mathrm{O_2(g) \rightleftharpoons 2\mathrm{SO_3(g)$ $K_c = 4.0 \times 10^{24}$ at $700 \mathrm{ K$ . If $0.10 \mathrm{ mol$ of $\mathrm{SO_2$ and $0.050 \mathrm{ mol$ of $\mathrm{O_2$ are placed in a $1.00 \mathrm{ L$ container, find the equilibrium concentrations.
Explain why increasing the concentration of a reactant in a reaction at equilibrium causes more product to form, but does not change the value of $K$ .
The decomposition of $\mathrm{HI$ is second order with a rate constant of $1.6 \times 10^{-3} \mathrm{ M^{-1}\mathrm{s^{-1}$ at $700 \mathrm{ K$ . If the initial concentration of $\mathrm{HI$ is $0.200 \mathrm{ M$ How long does it take for the concentration to decrease to $0.050 \mathrm{ M$ ?
A catalyst lowers the activation energy of a reaction from $85 \mathrm{ kJ/mol$ to $55 \mathrm{kJ/mol$ . Calculate the ratio of rate constants at $300 \mathrm{ K$ .
For $\mathrm{H_2(g) + \mathrm{I_2(g) \rightleftharpoons 2\mathrm{HI(g)$ at $448^{\circ}\mathrm{C$ , $K_c = 50.5$ . Calculate $K_p$ for this reaction at the same temperature.
Will a precipitate form when equal volumes of $0.0020 \mathrm{ M \mathrm{CaCl_2$ and $0.0010 \mathrm{ M \mathrm{Na_2\mathrm{SO_4$ are mixed? $K_{sp}(\mathrm{CaSO_4) = 2.4 \times 10^{-5}$ .
For a zero-order reaction $\mathrm{A \to \mathrm{products$ with $k = 0.0050 \mathrm{ M/s$ calculate the concentration of $\mathrm{A$ after $60 \mathrm{ s$ if $[\mathrm{A]_0 = 0.400 \mathrm{ M$ .
Calculate $K_c$ for the reaction $\mathrm{Fe^{3+}(aq) + \mathrm{SCN^-(aq) \rightleftharpoons \mathrm{FeSCN^{2+}(aq)$ if at equilibrium $[\mathrm{Fe^{3+}] = 0.0100 \mathrm{ M$ , $[\mathrm{SCN^-] = 0.0080 \mathrm{ M$ and $[\mathrm{FeSCN^{2+}] = 0.0020 \mathrm{ M$ .
Explain why the rate of a reaction approximately doubles for every $10^{\circ}\mathrm{C$ increase in temperature (the “rule of thumb”), and show that this corresponds to an activation energy of approximately $50 \mathrm{ kJ/mol$ using the Arrhenius equation.
For the reaction $\mathrm{N_2\mathrm{O_4(g) \rightleftharpoons 2\mathrm{NO_2(g)$ $K_c = 0.600$ at $340 \mathrm{ K$ . If $1.00 \mathrm{ atm$ of $\mathrm{N_2\mathrm{O_4$ is placed in a container at $340 \mathrm{ K$ Find the equilibrium partial pressures and the percentage dissociation.
Calculate the pH of a saturated solution of $\mathrm{Mg(OH)_2$ . $K_{sp} = 5.6 \times 10^{-12}$ .
A reaction has $\Delta H = +50 \mathrm{ kJ/mol$ . At $300 \mathrm{ K$ , $K = 0.10$ . Calculate $K$ at $400 \mathrm{ K$ using the van’t Hoff equation.

Practice Problems

Question 1: Reaction order determination from initial rates

For the reaction $\mathrm{A + \mathrm{B \to \mathrm{C$ The following initial rate data were Collected:

$[\mathrm{A]$ (M)	$[\mathrm{B]$ (M)	Initial Rate (M/s)
0.10	0.10	0.0020
0.20	0.10	0.0040
0.10	0.20	0.0080

Determine the rate law, the overall order, and the rate constant $k$ .

Answer

Comparing experiments 1 and 2: $[\mathrm{B]$ is constant, $[\mathrm{A]$ doubles, rate doubles. Rate is first order in A.

Comparing experiments 1 and 3: $[\mathrm{A]$ is constant, $[\mathrm{B]$ doubles, rate quadruples. Rate is second order in B.

Rate law: $\mathrm{Rate = k[\mathrm{A][\mathrm{B]^2$ .

Overall order: $1 + 2 = 3$ .

Using experiment 1: $0.0020 = k(0.10)(0.10)^2 = k(0.001)$ So $k = 0.0020 / 0.001 = 2.0 \mathrm{ M^{-2}s^{-1}}$ .

Question 2: Equilibrium calculation with ICE table

At $500 \mathrm{ K$ , $\mathrm{PCl_5(g) \rightleftharpoons \mathrm{PCl_3(g) + \mathrm{Cl_2(g)$ Has $K_p = 1.05$ . If $2.00 \mathrm{ atm$ of $\mathrm{PCl_5$ is placed in a flask and the system Reaches equilibrium, calculate the equilibrium partial pressures of all three gases and the Percentage dissociation of $\mathrm{PCl_5$ .

Answer

ICE table (pressures in atm):

	$\mathrm{PCl_5$	$\mathrm{PCl_3$	$\mathrm{Cl_2$
I	2.00	0	0
C	$-x$	$+x$	$+x$
E	$2.00 - x$	$x$	$x$

$K_p = \frac{P_{\mathrm{PCl_3} \cdot P_{\mathrm{Cl_2}}{P_{\mathrm{PCl_5}} = \frac{x \cdot x}{2.00 - x} = 1.05$

$x^2 = 1.05(2.00 - x) = 2.10 - 1.05x$

$x^2 + 1.05x - 2.10 = 0$

Using the quadratic formula: $x = \frac{-1.05 + \sqrt{1.1025 + 8.40}}{2} = \frac{-1.05 + 3.086}{2} = 1.018 \mathrm{ atm$ .

Equilibrium pressures: $P_{\mathrm{PCl_5} = 2.00 - 1.018 = 0.982 \mathrm{ atm$ $P_{\mathrm{PCl_3} = 1.018 \mathrm{ atm$ , $P_{\mathrm{Cl_2} = 1.018 \mathrm{ atm$ .

Percentage dissociation: $\frac{1.018}{2.00} \times 100 = 50.9\%$ .

Question 3: Le Chatelier's principle with pressure and temperature

For the exothermic reaction $\mathrm{N_2(g) + 3\mathrm{H_2(g) \rightleftharpoons 2\mathrm{NH_3(g)$ Predict the effect on the Equilibrium yield of $\mathrm{NH_3$ when (a) total pressure is increased, (b) temperature is Increased, (c) a catalyst is added, and (d) $\mathrm{Ar(g)$ is added at constant volume.

Answer

(a) Increasing total pressure shifts equilibrium toward the side with fewer moles of gas. Reactants: 4 mol gas; products: 2 mol gas. Equilibrium shifts right, increasing $\mathrm{NH_3$ yield.

(b) Increasing temperature favours the endothermic direction. Since the reaction is exothermic, Increasing temperature shifts equilibrium left, decreasing $\mathrm{NH_3$ yield.

(c) Adding a catalyst increases the rate of both forward and reverse reactions equally. It does not Shift the equilibrium position or change the yield. It only helps the system reach equilibrium Faster.

(d) Adding $\mathrm{Ar$ at constant volume increases the total pressure but does not change the Partial pressures of the reactants or products (since volume is constant and no new moles of Reactant/product are added). There is no shift in equilibrium.

Question 4: Arrhenius equation and activation energy

A reaction has a rate constant of $3.46 \times 10^{-5} \mathrm{ s^{-1}}$ at $298 \mathrm{ K$ and $4.87 \times 10^{-3} \mathrm{ s^{-1}}$ at $350 \mathrm{ K$ . Calculate the activation energy $E_a$ And the pre-exponential factor $A$ .

Answer

Using the two-point form of the Arrhenius equation:

$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

$\ln\frac{4.87 \times 10^{-3}}{3.46 \times 10^{-5}} = \frac{E_a}{8.314}\left(\frac{1}{298} - \frac{1}{350}\right)$

$\ln(140.8) = \frac{E_a}{8.314}(0.003356 - 0.002857)$

$4.947 = \frac{E_a}{8.314}(0.000499)$

$E_a = \frac{4.947 \times 8.314}{0.000499} = \frac{41.13}{0.000499} = 82,400 \mathrm{ J/mol = 82.4 \mathrm{ kJ/mol$

For the pre-exponential factor $A$ Using $k = Ae^{-E_a/RT}$ at $298 \mathrm{ K$ :

$3.46 \times 10^{-5} = A \cdot e^{-82400/(8.314 \times 298)} = A \cdot e^{-33.28}$

$A = \frac{3.46 \times 10^{-5}}{3.62 \times 10^{-15}} = 9.56 \times 10^{9} \mathrm{ s^{-1}}$

Question 5: Solubility product and common ion effect

The $K_{sp}$ of $\mathrm{PbCl_2$ is $1.7 \times 10^{-5}$ at $25^\circ\mathrm{C$ . Calculate (a) the Molar solubility of $\mathrm{PbCl_2$ in pure water, and (b) the molar solubility in a $0.10 \mathrm{ M$ $\mathrm{NaCl$ solution.

Answer

(a) In pure water: Let $s$ = molar solubility.

$\mathrm{PbCl_2(s) \rightleftharpoons \mathrm{Pb^{2+}(aq) + 2\mathrm{Cl^-(aq)$

$K_{sp} = [\mathrm{Pb^{2+}][\mathrm{Cl^-]^2 = s \times (2s)^2 = 4s^3 = 1.7 \times 10^{-5}$

$s^3 = 4.25 \times 10^{-6}$ So $s = 1.62 \times 10^{-2} \mathrm{ M$ .

(b) In $0.10 \mathrm{ M$ $\mathrm{NaCl$ : $[\mathrm{Cl^-] = 0.10 \mathrm{ M$ initially.

$K_{sp} = [\mathrm{Pb^{2+}][\mathrm{Cl^-]^2 = s \times (0.10 + 2s)^2$

Assuming $2s \ll 0.10$ : $1.7 \times 10^{-5} = s \times (0.10)^2 = 0.01s$

$s = 1.7 \times 10^{-3} \mathrm{ M$ .

The common ion effect reduces the solubility from $1.62 \times 10^{-2} \mathrm{ M$ to $1.7 \times 10^{-3} \mathrm{ M$ Approximately a 10-fold decrease.

Worked Examples

Example 1: Mole calculation

Calculate the number of moles in $12.0\,\text{g}$ of $\text{NaOH}$ ( $M_r = 40.0$ ).

Solution:

$n = \frac{m}{M_r} = \frac{12.0}{40.0} = 0.300\,\text{mol}$

Example 2: Reacting masses

$\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$

What mass of $\text{CaCl}_2$ is produced from $10.0\,\text{g}$ of $\text{CaCO}_3$ ? ( $M_r[\text{CaCO}_3] = 100$ , $M_r[\text{CaCl}_2] = 111$ )

Solution:

$n(\text{CaCO}_3) = \frac{10.0}{100} = 0.100\,\text{mol}$

From the equation, ratio is $1:1$ , so $n(\text{CaCl}_2) = 0.100\,\text{mol}$ .

$m(\text{CaCl}_2) = 0.100 \times 111 = 11.1\,\text{g}$

Intuition

Chemistry studies how atoms combine and react to form everything around us. The periodic table organises elements by their properties, chemical bonds hold molecules together, and reactions transform one substance into another. From the air we breathe to the food we eat, chemistry explains the material basis of life and the principles behind countless technologies.

Cross-References

Mark this page as reviewed