Enthalpy ($H$): The heat content of a system at constant pressure.

Standard enthalpy change of reaction ($\Delta H_r^\circ$): The enthalpy change when molar Quantities of reactants as stated in the equation react under standard conditions (298 K, 100 kPa).

Exothermic: $\Delta H < 0$ (energy released to surroundings).

Endothermic: $\Delta H > 0$ (energy absorbed from surroundings).

Hess’s Law states that the enthalpy change of a reaction is independent of the route taken, provided The initial and final conditions are the same.

$\Delta H_1 = \Delta H_2 + \Delta H_3$

Proof of Hess’s Law:

Enthalpy is a state function, meaning only on the initial and final states, not on the Pathway. Since enthalpy is defined as $H = U + pV$ and $U$ (internal energy) and $pV$ are both state Functions, $H$ must also be a state function. Therefore, any path between the same initial and final States must give the same $\Delta H$.

Worked Example 1: Calculate $\Delta H_f^\circ$ for \mathrm{CH_4 given:

\mathrm{C(s) + \mathrm{O_2\mathrm{(g) \to \mathrm{CO_2\mathrm{(g) \quad \Delta H = -393.5 \mathrm{ kJ/mol

\mathrm{H_2\mathrm{(g) + \tfrac{1}{2}\mathrm{O_2\mathrm{(g) \to \mathrm{H_2\mathrm{O(l) \quad \Delta H = -285.8 \mathrm{ kJ/mol

\mathrm{CH_4\mathrm{(g) + 2\mathrm{O_2\mathrm{(g) \to \mathrm{CO_2\mathrm{(g) + 2\mathrm{H_2\mathrm{O(l) \quad \Delta H = -890.3 \mathrm{ kJ/mol

Using Hess’s Law (elements $\to$ products via two routes):

\Delta H_f(\mathrm{CH_4) + (-890.3) = -393.5 + 2(-285.8)

\Delta H_f(\mathrm{CH_4) = -393.5 - 571.6 + 890.3 = -74.8 \mathrm{ kJ/mol

Worked Example 2: Calculate $\Delta H_f^\circ$ for \mathrm{CS_2 given:

\mathrm{C(s) + \mathrm{O_2\mathrm{(g) \to \mathrm{CO_2\mathrm{(g) \quad \Delta H = -393.5 \mathrm{ kJ/mol

\mathrm{S(s) + \mathrm{O_2\mathrm{(g) \to \mathrm{SO_2\mathrm{(g) \quad \Delta H = -296.8 \mathrm{ kJ/mol

\mathrm{CS_2\mathrm{(l) + 3\mathrm{O_2\mathrm{(g) \to \mathrm{CO_2\mathrm{(g) + 2\mathrm{SO_2\mathrm{(g) \quad \Delta H = -1075 \mathrm{ kJ/mol

Route 1: \mathrm{C + 2\mathrm{S \to \mathrm{CS_2 (direct, $\Delta H_f$) Route 2: \mathrm{C + \mathrm{O_2 \to \mathrm{CO_2 and 2\mathrm{S + 2\mathrm{O_2 \to 2\mathrm{SO_2Then \mathrm{CO_2 + 2\mathrm{SO_2 \to \mathrm{CS_2 + 3\mathrm{O_2 (reverse the combustion)

\Delta H_f = -393.5 + 2(-296.8) - (-1075) = -393.5 - 593.6 + 1075 = 87.9 \mathrm{ kJ/mol

Enthalpy of combustion:

$q = mc\Delta T$

Where $m$ is mass of water, $c$ is specific heat capacity (4.18 \mathrm{ J g^{-1}\mathrm{K^{-1}), And $\Delta T$ is temperature change.

Worked Example 3: When 1.50 \mathrm{ g of ethanol is burned, it raises the temperature of 200 \mathrm{ g of water by $14.2°C$. Calculate the enthalpy of combustion.

q = 200 \times 4.18 \times 14.2 = 11871.2 \mathrm{ J = 11.87 \mathrm{ kJ

n(\mathrm{ethanol) = \frac{1.50}{46.07} = 0.03256 \mathrm{ mol

\Delta H_c = -\frac{11.87}{0.03256} = -364.7 \mathrm{ kJ/mol

(The negative sign indicates exothermic.)

Worked Example 4: 0.80 \mathrm{ g of ethanol ($M_r = 46$) raised the temperature of 150 \mathrm{ g of water by $10.5°C$. Calculate $\Delta H_c$ and suggest why this differs from the Literature value of -1367 \mathrm{ kJ/mol.

q = 150 \times 4.18 \times 10.5 = 6583.5 \mathrm{ J = 6.58 \mathrm{ kJ

n = \frac{0.80}{46} = 0.0174 \mathrm{ mol

\Delta H_c = -\frac{6.58}{0.0174} = -378 \mathrm{ kJ/mol

This is much less exothermic than the literature value because of:

• Heat loss to the surroundings (calorimeter is not perfectly insulated)
• Incomplete combustion of ethanol
• Not all heat was transferred to the water (some heated the calorimeter itself)

Worked Example 5: 25.0 \mathrm{ cm^3 of 1.0 \mathrm{ M HCl is mixed with 25.0 \mathrm{ cm^3 Of 1.0 \mathrm{ M NaOH. The temperature rises by $6.8°C$. Calculate the enthalpy of neutralisation.

q = 50.0 \times 4.18 \times 6.8 = 1421.2 \mathrm{ J = 1.42 \mathrm{ kJ

n = 1.0 \times 0.0250 = 0.0250 \mathrm{ mol

\Delta H_{\mathrm{neut} = -\frac{1.42}{0.0250} = -56.8 \mathrm{ kJ/mol

In a reversible reaction, when the rate of the forward reaction equals the rate of the reverse Reaction, the system is at dynamic equilibrium.

$aA + bB \rightleftharpoons cC + dD$

Conditions for dynamic equilibrium:

1. The reaction must be reversible
2. The system must be closed (no matter can enter or leave)
3. The temperature must be constant
4. Macroscopic properties (concentration, colour, pressure) remain constant

If a system at equilibrium is subjected to a change, the system adjusts to oppose that change.

Example: For the Haber process: \mathrm{N_2 + 3\mathrm{H_2 \rightleftharpoons 2\mathrm{NH_3 \quad \Delta H = -92 \mathrm{ kJ/mol.

• High pressure favours \mathrm{NH_3 (4 moles $\to$ 2 moles of gas)
• Low temperature favours \mathrm{NH_3 (exothermic), but slow rate
• Compromise: moderate temperature (~450°C), high pressure (~200 atm), iron catalyst

Worked Example 6: Explain the effect of increasing pressure on the equilibrium \mathrm{N_2\mathrm{O_4\mathrm{(g) \rightleftharpoons 2\mathrm{NO_2\mathrm{(g).

The forward reaction produces 2 moles of gas from 1 mole. Increasing pressure favours the side with Fewer gas moles, so the equilibrium shifts to the left, favouring \mathrm{N_2\mathrm{O_4. The Mixture becomes paler (less brown \mathrm{NO_2 gas).

For the reaction $aA + bB \rightleftharpoons cC + dD$:

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

Where square brackets denote equilibrium concentrations in \mathrm{mol/L.

Key points about $K_c$:

• $K_c$ is only affected by temperature, not by concentration, pressure, or catalysts
• A large $K_c$ ($\gt 10^3$) means products are favoured
• A small $K_c$ ($\lt 10^{-3}$) means reactants are favoured
• Pure solids and pure liquids are NOT included in the $K_c$ expression

Worked Example 7: For \mathrm{H_2 + \mathrm{I_2 \rightleftharpoons 2\mathrm{HIAt equilibrium The concentrations are [\mathrm{H_2] = 0.22, [\mathrm{I_2] = 0.22 [\mathrm{HI] = 1.56 \mathrm{ mol/L. Find $K_c$.

K_c = \frac{[\mathrm{HI]^2}{[\mathrm{H_2][\mathrm{I_2]} = \frac{(1.56)^2}{(0.22)(0.22)} = \frac{2.4336}{0.0484} = 50.3

Worked Example 8: 2.0 \mathrm{ mol of \mathrm{SO_2 and 1.0 \mathrm{ mol of \mathrm{O_2 are Placed in a 1.0 \mathrm{ L flask at equilibrium. If 1.4 \mathrm{ mol of \mathrm{SO_3 forms, find $K_c$ for \mathrm{2SO_2 + \mathrm{O_2 \rightleftharpoons 2\mathrm{SO_3.

K_c = \frac{[\mathrm{SO_3]^2}{[\mathrm{SO_2]^2[\mathrm{O_2]} = \frac{(1.4)^2}{(0.6)^2(0.3)} = \frac{1.96}{0.108} = 18.15

Worked Example 9: 1.0 \mathrm{ mol of \mathrm{PCl_5 is placed in a 5.0 \mathrm{ L container. At equilibrium, 0.3 \mathrm{ mol has dissociated: \mathrm{PCl_5 \rightleftharpoons \mathrm{PCl_3 + \mathrm{Cl_2. Find $K_c$.

Concentrations (divide by $V = 5.0$ L): [\mathrm{PCl_5] = 0.14, [\mathrm{PCl_3] = 0.06 [\mathrm{Cl_2] = 0.06 \mathrm{ mol/L.

K_c = \frac{[\mathrm{PCl_3][\mathrm{Cl_2]}{[\mathrm{PCl_5]} = \frac{(0.06)(0.06)}{0.14} = \frac{0.0036}{0.14} = 0.0257

$\Delta G^\circ = -RT\ln K$

Where R = 8.314 \mathrm{ J mol^{-1}\mathrm{K^{-1}, $T$ is temperature in Kelvin.

Derivation:

Starting from the thermodynamic relationship $\Delta G = \Delta G^\circ + RT\ln Q$ (where $Q$ is the Reaction quotient), at equilibrium $\Delta G = 0$ and $Q = K$Giving:

$0 = \Delta G^\circ + RT\ln K \implies \Delta G^\circ = -RT\ln K$

• $\Delta G^\circ < 0$: $K > 1$Reaction favours products
• $\Delta G^\circ > 0$: $K < 1$Reaction favours reactants
• $\Delta G^\circ = 0$: $K = 1$System at equilibrium

Worked Example 10: Calculate $K$ at 298 \mathrm{ K for a reaction with \Delta G^\circ = -15.2 \mathrm{ kJ/mol.

$\Delta G^\circ = -RT\ln K \implies -15200 = -8.314 \times 298 \times \ln K$

$\ln K = \frac{15200}{8.314 \times 298} = \frac{15200}{2477.6} = 6.135$

$K = e^{6.135} = 461$

$\ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

This relates the equilibrium constant at two different temperatures.

Derivation:

Starting from $\Delta G^\circ = -RT\ln K$ and $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$:

$-RT\ln K = \Delta H^\circ - T\Delta S^\circ$

$\ln K = -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R}$

Assuming $\Delta H^\circ$ and $\Delta S^\circ$ are temperature-independent over the range of Interest, differentiating with respect to $T$:

$\frac{d\ln K}{dT} = \frac{\Delta H^\circ}{RT^2}$

Integrating from $T_1$ to $T_2$:

$\ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

For a reaction to occur:

1. Particles must collide
2. They must collide with sufficient energy (greater than or equal to the activation energy, $E_a$)
3. They must collide with the correct orientation

At any temperature, molecules have a range of kinetic energies. The Maxwell-Boltzmann distribution Shows that:

• Most molecules have energies close to the average
• Few molecules have very low or very high energies
• Increasing temperature shifts the curve to the right and increases the proportion of molecules with energy above $E_a$

Effect of temperature: A small increase in temperature significantly increases the fraction of Molecules with energy above $E_a$Because the distribution is exponential. A 10^\circ\mathrm{C rise Doubles the rate.

Effect of a catalyst: A catalyst provides an alternative reaction pathway with a lower Activation energy. It does not change the distribution of molecular energies but lowers the Threshold, meaning more molecules have sufficient energy to react.

For a reaction aA + bB \to \mathrm{products:

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

Where $m$ and $n$ are the orders of reaction with respect to A and B, and $k$ is the rate Constant.

The overall order is $m + n$.

Units of $k$: Depend on the overall order.

Worked Example 11: For the reaction \mathrm{A + 2\mathrm{B \to \mathrm{CThe rate equation is \mathrm{Rate = k[\mathrm{A][\mathrm{B]^2. If [\mathrm{A] doubles and [\mathrm{B] triples, by what Factor does the rate change?

New rate = k(2[\mathrm{A])(3[\mathrm{B])^2 = k \times 2[\mathrm{A] \times 9[\mathrm{B]^2 = 18 \times k[\mathrm{A][\mathrm{B]^2.

The rate increases by a factor of 18.

Method of initial rates: Vary the concentration of one reactant while keeping others constant. Measure the initial rate and determine the order with respect to each reactant.

Worked Example 12: Given the following data for the reaction \mathrm{A + \mathrm{B \to \mathrm{products:

From experiments 1 and 2: doubling [\mathrm{A] doubles the rate, so order with respect to \mathrm{A is 1. From experiments 1 and 3: doubling [\mathrm{B] quadruples the rate, so order with Respect to \mathrm{B is 2.

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

k = \frac{\mathrm{Rate}{[\mathrm{A][\mathrm{B]^2} = \frac{1.2 \times 10^{-3}}{(0.10)(0.10)^2} = \frac{1.2 \times 10^{-3}}{0.001} = 1.2 \mathrm{ L^2\mathrm{mol^{-2}\mathrm{s^{-1}

$k = Ae^{-E_a/RT}$

$\ln k = \ln A - \frac{E_a}{RT}$

A plot of $\ln k$ vs. $1/T$ gives a straight line with gradient $-E_a/R$ and y-intercept $\ln A$.

Derivation of the Arrhenius equation:

The Arrhenius equation arises from the observation that the rate constant depends exponentially on Temperature. The factor $e^{-E_a/RT}$ represents the fraction of collisions with energy $\geq E_a$. The pre-exponential factor $A$ (frequency factor) accounts for the frequency of collisions and the Orientation factor.

Worked Example 13: The rate constant of a reaction at 300 \mathrm{ K is 2.5 \times 10^{-3} \mathrm{ s^{-1} and at 350 \mathrm{ K is 4.2 \times 10^{-2} \mathrm{ s^{-1}. Find the activation energy.

$\ln\left(\frac{4.2 \times 10^{-2}}{2.5 \times 10^{-3}}\right) = -\frac{E_a}{8.314}\left(\frac{1}{350} - \frac{1}{300}\right)$

$\ln(16.8) = -\frac{E_a}{8.314}\left(0.002857 - 0.003333\right)$

$2.821 = -\frac{E_a}{8.314}(-0.000476)$

$2.821 = \frac{E_a \times 0.000476}{8.314}$

E_a = \frac{2.821 \times 8.314}{0.000476} = 49239 \mathrm{ J/mol \approx 49.2 \mathrm{ kJ/mol

Worked Example 14: The activation energy of a reaction is 75 \mathrm{ kJ/mol. If the rate Constant at 300 \mathrm{ K is 3.0 \times 10^{-4} \mathrm{ s^{-1}Find the rate constant at 350 \mathrm{ K.

$\ln\left(\frac{k_2}{3.0 \times 10^{-4}}\right) = -\frac{75000}{8.314}\left(\frac{1}{350} - \frac{1}{300}\right)$

$\ln\left(\frac{k_2}{3.0 \times 10^{-4}}\right) = -9020 \times (-0.000476) = 4.294$

$\frac{k_2}{3.0 \times 10^{-4}} = e^{4.294} = 73.2$

k_2 = 73.2 \times 3.0 \times 10^{-4} = 2.20 \times 10^{-2} \mathrm{ s^{-1}

A catalyst provides an alternative reaction pathway with a lower activation energy. It does this by:

1. Forming an intermediate with the reactant(s)
2. The intermediate then decomposes to give products and regenerates the catalyst

Example: In the Haber process, iron catalyses the reaction by adsorbing \mathrm{N_2 and \mathrm{H_2 onto its surface. The bonds in \mathrm{N_2 are weakened, allowing hydrogen atoms to Add step-by-step at lower energy than the uncatalysed reaction.

For an exothermic reaction:

Energy
  ^
  |         ___
  |        /   \        Products (lower energy)
  |  ___  /     \___
  | /   \/           \___   Reactants
  |/   Ea(catalysed)      \___
  |        Ea(uncatalysed)
  +---------------------------------> Reaction progress

The catalysed pathway has a lower peak ($E_a$ reduced) but the same overall $\Delta H$.

1. Hess’s Law sign conventions: When reversing a reaction, reverse the sign of $\Delta H$.

2. Calorimetry heat losses: Real experiments lose heat to surroundings, so calculated values are less exothermic than literature values.

3. Le Chatelier’s Principle: A catalyst does NOT shift the equilibrium position; it only speeds up reaching equilibrium.

4. Units of $K_c$: Always include units, which depend on the stoichiometry of the reaction.

5. Rate order vs. Stoichiometric coefficient: The order of reaction is determined experimentally, not from the balanced equation.

6. Exothermic vs. Endothermic: Remember that $\Delta H$ refers to the system. Exothermic means the system loses energy ($\Delta H < 0$).

7. Equilibrium concentrations: Use the ICE table method (Initial, Change, Equilibrium) consistently to avoid errors.

1. Calculate the standard enthalpy of formation of \mathrm{CS_2 given:

• \mathrm{C(s) + \mathrm{O_2\mathrm{(g) \to \mathrm{CO_2\mathrm{(g) \Delta H = -393.5 \mathrm{ kJ/mol
• \mathrm{S(s) + \mathrm{O_2\mathrm{(g) \to \mathrm{SO_2\mathrm{(g) \Delta H = -296.8 \mathrm{ kJ/mol
• \mathrm{CS_2\mathrm{(l) + 3\mathrm{O_2\mathrm{(g) \to \mathrm{CO_2\mathrm{(g) + 2\mathrm{SO_2\mathrm{(g) \Delta H = -1075 \mathrm{ kJ/mol

2. Explain the effect of increasing pressure on the equilibrium \mathrm{N_2\mathrm{O_4\mathrm{(g) \rightleftharpoons 2\mathrm{NO_2\mathrm{(g).

3. For the reaction \mathrm{A + 2\mathrm{B \to \mathrm{CThe rate equation is \mathrm{Rate = k[\mathrm{A][\mathrm{B]^2. If [\mathrm{A] doubles and [\mathrm{B] triples, by what factor does the rate change?

4. The activation energy of a reaction is 75 \mathrm{ kJ/mol. If the rate constant at 300 \mathrm{ K is 3.0 \times 10^{-4} \mathrm{ s^{-1}Find the rate constant at 350 \mathrm{ K.

5. 1.0 \mathrm{ mol of \mathrm{PCl_5 is placed in a 5.0 \mathrm{ L container. At equilibrium, 0.3 \mathrm{ mol has dissociated: \mathrm{PCl_5 \rightleftharpoons \mathrm{PCl_3 + \mathrm{Cl_2. Find $K_c$.

6. Explain why a catalyst increases the rate of a reaction without being consumed.

7. In a calorimetry experiment, 0.80 \mathrm{ g of ethanol ($M_r = 46$) raised the temperature of 150 \mathrm{ g of water by $10.5°C$. Calculate $\Delta H_c$ and suggest why this value differs from the literature value of -1367 \mathrm{ kJ/mol.

8. Given $\Delta G^\circ = -RT\ln K$Calculate $K$ at 298 \mathrm{ K for a reaction with \Delta G^\circ = -15.2 \mathrm{ kJ/mol.

9. Using the following data, determine the rate equation and the value of $k$:

10. Sketch an energy profile diagram for an endothermic reaction, showing the effect of adding a catalyst. Label the activation energies, $\Delta H$And the transition state.

11. For the equilibrium \mathrm{CO(g) + \mathrm{H_2\mathrm{O(g) \rightleftharpoons \mathrm{CO_2\mathrm{(g) + \mathrm{H_2\mathrm{(g) \Delta H = -41 \mathrm{ kJ/molExplain how increasing temperature affects the yield of \mathrm{CO_2 and the value of $K_c$.

12. Calculate the enthalpy of neutralisation when 30.0 \mathrm{ cm^3 of 0.50 \mathrm{ M HCl is mixed with 30.0 \mathrm{ cm^3 of 0.50 \mathrm{ M NaOH, producing a temperature rise of $3.2°C$.

Used for reactions occurring in solution. The calorimeter itself also absorbs heat:

q_{\mathrm{total} = (m_{\mathrm{water}c_{\mathrm{water} + C_{\mathrm{cal})\Delta T

Where C_{\mathrm{cal} is the heat capacity of the calorimeter (in J/K).

Worked Example 15: A calorimeter has heat capacity 45 \mathrm{ J/K. When 50 \mathrm{ cm^3 of 1.0 \mathrm{ M \mathrm{HCl is mixed with 50 \mathrm{ cm^3 of 1.0 \mathrm{ M \mathrm{NaOH in the Calorimeter, the temperature rises by $6.5°C$. Calculate the enthalpy of neutralisation.

q_{\mathrm{total} = (100 \times 4.18 + 45) \times 6.5 = (418 + 45) \times 6.5 = 463 \times 6.5 = 3009.5 \mathrm{ J = 3.01 \mathrm{ kJ

n = 1.0 \times 0.050 = 0.050 \mathrm{ mol

\Delta H_{\mathrm{neut} = -\frac{3.01}{0.050} = -60.2 \mathrm{ kJ/mol

Used for combustion reactions. The sample is ignited in a sealed container (bomb) surrounded by Water. The bomb calorimeter is designed to operate at constant volume, so the heat measured is $\Delta U$ (internal energy change), not $\Delta H$. The correction is:

$\Delta H = \Delta U + \Delta n_g RT$

Where $\Delta n_g$ is the change in moles of gas.

Worked Example 16: For the reaction \mathrm{N_2\mathrm{O_4\mathrm{(g) \rightleftharpoons 2\mathrm{NO_2\mathrm{(g) \Delta H^\circ = +57.2 \mathrm{ kJ/mol and $K_c = 0.115$ at 298 \mathrm{ K. Find $K_c$ at 350 \mathrm{ K.

Using the van’t Hoff equation:

$\ln\left(\frac{K_2}{0.115}\right) = -\frac{57200}{8.314}\left(\frac{1}{350} - \frac{1}{298}\right)$

$= -6879 \times (0.002857 - 0.003356) = -6879 \times (-0.000499) = 3.432$

$\frac{K_2}{0.115} = e^{3.432} = 30.94$

$K_2 = 30.94 \times 0.115 = 3.56$

The equilibrium constant increases significantly with temperature, confirming that the forward Reaction is endothermic (Le Chatelier’s principle).

For a multi-step reaction, the overall rate is determined by the rate-determining step (slowest Step).

Worked Example 17: A reaction proceeds by the following mechanism:

Step 1 (slow): \mathrm{A + \mathrm{B \to \mathrm{C

Step 2 (fast): \mathrm{C + \mathrm{D \to \mathrm{E

The rate equation is determined by the slow step: \mathrm{Rate = k[\mathrm{A][\mathrm{B].

Note that \mathrm{D does not appear in the rate equation because it is involved only in the fast Step.

Worked Example 18: For the reaction \mathrm{2NO_2 + \mathrm{F_2 \to 2\mathrm{NO_2\mathrm{FThe Proposed mechanism is:

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 equation is: \mathrm{Rate = k[\mathrm{NO_2][\mathrm{F_2].

This is first order with respect to \mathrm{NO_2 and first order with respect to \mathrm{F_2Even Though the overall balanced equation has coefficient 2 for \mathrm{NO_2. This confirms that the Rate order cannot be predicted from the stoichiometry.

13. Using Hess’s Law, calculate $\Delta H_r^\circ$ for: \mathrm{3C(s) + 4\mathrm{H_2\mathrm{(g) \to \mathrm{C_3\mathrm{H_8\mathrm{(g) given \Delta H_c^\circ(\mathrm{C_3\mathrm{H_8) = -2220 \mathrm{ kJ/mol \Delta H_c^\circ(\mathrm{C) = -393.5 \mathrm{ kJ/mol \Delta H_c^\circ(\mathrm{H_2) = -285.8 \mathrm{ kJ/mol.

14. Explain why the enthalpy of neutralisation for strong acid-strong base reactions is approximately constant (~-57 kJ/mol), but for weak acid-strong base reactions it is less exothermic.

15. For a reaction with \Delta H^\circ = -92 \mathrm{ kJ/molExplain qualitatively how $K_c$ changes as temperature increases. Would the equilibrium yield of products increase or decrease?

16. In a bomb calorimeter, 1.00 \mathrm{ g of glucose (\mathrm{C_6\mathrm{H_{12}\mathrm{O_6) is burned. The temperature of 500 \mathrm{ g of water rises by $7.48°C$. The calorimeter heat capacity is 620 \mathrm{ J/K. Calculate the enthalpy of combustion per mole of glucose.

17. The following data were obtained for the reaction \mathrm{A + \mathrm{B \to \mathrm{C at 25°C:

Determine the rate equation, the value of $k$ with units, and the overall order.

18. Draw and label a fully annotated Maxwell-Boltzmann distribution curve at two different temperatures, showing the activation energy and explaining why a small temperature increase has a large effect on the reaction rate.

Entropy is a measure of disorder or randomness in a system.

Factors that increase entropy:

• Increasing the number of particles (more ways to arrange them)
• Changing from solid to liquid to gas
• Increasing temperature (more kinetic energy, more microstates)
• Dissolving a solid in a solvent

Standard entropy change:

\Delta S^\circ = \sum S^\circ(\mathrm{products) - \sum S^\circ(\mathrm{reactants)

$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$

Worked Example 19: For the reaction \mathrm{CaCO_3\mathrm{(s) \to \mathrm{CaO(s) + \mathrm{CO_2\mathrm{(g) \Delta H^\circ = +178 \mathrm{ kJ/mol and \Delta S^\circ = +160 \mathrm{ J mol^{-1}\mathrm{K^{-1}. Find the minimum temperature at which the reaction becomes spontaneous.

The reaction is spontaneous when $\Delta G^\circ < 0$:

$0 = \Delta H^\circ - T\Delta S^\circ$

T = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{178000}{160} = 1112.5 \mathrm{ K \approx 840°C

This is the decomposition temperature of limestone, consistent with industrial practice.

Worked Example 20: For the dissolution of \mathrm{NH_4\mathrm{NO_3\mathrm{(s) in water: \Delta H^\circ = +25.7 \mathrm{ kJ/mol, \Delta S^\circ = +108 \mathrm{ J mol^{-1}\mathrm{K^{-1}. Explain why this dissolution is spontaneous at room temperature.

\Delta G^\circ = 25700 - 298 \times 108 = 25700 - 32184 = -6484 \mathrm{ J/mol

Since $\Delta G^\circ < 0$The dissolution is spontaneous despite being endothermic. The driving Force is the large increase in entropy (solid $\to$ aqueous ions).

Example 1: Conservation of energy

A $0.50\,\text{kg}$ ball is dropped from a height of $20\,\text{m}$. Calculate its speed just before it hits the ground (ignore air resistance).

Solution:

Using conservation of energy: $mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 20} = \sqrt{392.4} \approx 19.8\,\text{m\,s}^{-1}$