Thermodynamics

Energy cannot be created or destroyed. The change in internal energy of a system is:

$$\Delta U = q + w$$

Where $q$ is heat and $w$ is work. The first law is a statement of energy conservation: any change in the internal energy of a system must be accounted for by heat flow and work done.

• $q \gt 0$: heat absorbed by the system (endothermic)
• $q \lt 0$: heat released by the system (exothermic)
• $w \gt 0$: work done on the system (compression)
• $w \lt 0$: work done by the system (expansion)

This sign convention is used by chemists (IUPAC convention). Some physics texts use the opposite Sign for work.

For a gas expanding/contracting against constant external pressure:

$$W = -P\Delta V$$

The negative sign reflects the chemistry convention: when a gas expands ($\Delta V \gt 0$), it does Work on the surroundings ($w \lt 0$).

At constant pressure, $w = -P\Delta V$ So:

$$\Delta U = q_P - P\Delta V \implies q_P = \Delta U + P\Delta V = \Delta H$$

Enthalpy ($H$) is defined as $H = U + PV$. At constant pressure, $\Delta H = q_P$. Enthalpy is Convenient because most chemical reactions occur at constant (atmospheric) pressure.

Starting from the first law at constant pressure:

$$\Delta U = q_P + w = q_P - P\Delta V$$ $$q_P = \Delta U + P\Delta V$$ $$q_P = (U_2 - U_1) + P(V_2 - V_1) = (U_2 + PV_2) - (U_1 + PV_1) = H_2 - H_1 = \Delta H$$

This derivation shows that enthalpy change equals heat at constant pressure because the $P\Delta V$ Work term is absorbed into the enthalpy definition.

The enthalpy change when 1 mole of a compound forms from its elements in their standard states.

$$\Delta H_{\mathrm{rxn}^\circ} = \sum n\Delta H_f^\circ(\mathrm{products)} - \sum m\Delta H_f^\circ(\mathrm{reactants)}$$

Standard conditions: $1 \mathrm{ atm}$, $298 \mathrm{ K}$ ($25^\circ\mathrm{C}$), pure substances in their most stable form. By convention, $\Delta H_f^\circ = 0$ for elements in their standard state.

The standard state of an element is its most stable form at $1 \mathrm{ atm}$ and $25^\circ\mathrm{C}$: e.g., $\mathrm{O_2(g)}$ Not $\mathrm{O_3(g)}$ or $\mathrm{O_2(l)}$; $\mathrm{C(graphite)}$ Not $\mathrm{C(diamond)}$.

The total enthalpy change for a reaction is the same regardless of the pathway. If a reaction can be written as the sum of several steps:

$$\Delta H_{\mathrm{total}} = \Delta H_1 + \Delta H_2 + \cdots$$

Hess’s law is a direct consequence of enthalpy being a state function: only on the Initial and final states, not on the path between them. This allows us to calculate enthalpy changes That cannot be measured directly.

Calculate $\Delta H^\circ$ for the reaction $\mathrm{C(\mathrm{graphite)}} + 2\mathrm{H_2(g)} \to \mathrm{CH_4(g)}$ using the following data:

$\mathrm{C(\mathrm{graphite)}} + \mathrm{O_2(g)} \to \mathrm{CO_2(g)}$ $\Delta H_1 = -393.5 \mathrm{ kJ/mol}$

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

$\mathrm{CH_4(g)} + 2\mathrm{O_2(g)} \to \mathrm{CO_2(g)} + 2\mathrm{H_2\mathrm{O(l)}}$ $\Delta H_3 = -890.3 \mathrm{ kJ/mol}$

Using Hess’s law: $\Delta H_f(\mathrm{CH_4)} = \Delta H_1 + 2\Delta H_2 - \Delta H_3$

$$= -393.5 + 2(-285.8) - (-890.3) = -393.5 - 571.6 + 890.3 = -74.8 \mathrm{ kJ/mol}$$

Calculate $\Delta H^\circ$ for $\mathrm{C(s)} + 2\mathrm{H_2(g)} + \frac{1}{2}\mathrm{O_2(g)} \to \mathrm{CH_3\mathrm{OH(l)}}$.

Given: $\mathrm{C(s)} + \mathrm{O_2(g)} \to \mathrm{CO_2(g)}$, $\Delta H = -393.5 \mathrm{ kJ/mol}$ $\mathrm{H_2(g)} + \frac{1}{2}\mathrm{O_2(g)} \to \mathrm{H_2\mathrm{O(l)}}$ $\Delta H = -285.8 \mathrm{ kJ/mol}$ $\mathrm{CH_3\mathrm{OH(l)}} + \frac{3}{2}\mathrm{O_2(g)} \to \mathrm{CO_2(g)} + 2\mathrm{H_2\mathrm{O(l)}}$ $\Delta H = -726.4 \mathrm{ kJ/mol}$

Target: $\mathrm{C(s)} + 2\mathrm{H_2(g)} + \frac{1}{2}\mathrm{O_2(g)} \to \mathrm{CH_3\mathrm{OH(l)}}$

Reverse equation 3 and add equations 1 and 2:

$\mathrm{C(s)} + \mathrm{O_2(g)} \to \mathrm{CO_2(g)}$, $\Delta H = -393.5$

$2\mathrm{H_2(g)} + \mathrm{O_2(g)} \to 2\mathrm{H_2\mathrm{O(l)}}$, $\Delta H = 2(-285.8) = -571.6$

$\mathrm{CO_2(g)} + 2\mathrm{H_2\mathrm{O(l)}} \to \mathrm{CH_3\mathrm{OH(l)}} + \frac{3}{2}\mathrm{O_2(g)}$ $\Delta H = +726.4$

Sum: $\Delta H = -393.5 - 571.6 + 726.4 = -238.7 \mathrm{ kJ/mol}$

Estimate $\Delta H$ for $\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \to 2\mathrm{NH_3(g)}$ using bond Enthalpies.

Bonds broken: $1 \mathrm{ N\equiv\mathrm{N (945)}} + 3 \mathrm{ H}-H (436) = 945 + 1308 = 2253 \mathrm{ kJ/mol}$

Bonds formed: $6 \mathrm{ N}-H (391) = 2346 \mathrm{ kJ/mol}$

$$\Delta H \approx 2253 - 2346 = -93 \mathrm{ kJ/mol}$$

(The exact value is $-92.2 \mathrm{ kJ/mol}$ Showing that bond enthalpies give a good Approximation.)

For the reaction $\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \to 2\mathrm{NH_3(g)}$ at $298 \mathrm{ K}$:

$\Delta H^\circ = -92.2 \mathrm{ kJ/mol}$ $\Delta S^\circ = -198.8 \mathrm{ J/(mol\cdot\mathrm{K)}}$.

Calculate $\Delta G^\circ$ and $K$.

$$\Delta G^\circ = -92200 - 298(-198.8) = -92200 + 59242 = -32958 \mathrm{ J/mol} = -33.0 \mathrm{ kJ/mol}$$ $$K = e^{-\Delta G^\circ/(RT)} = e^{32958/(8.314 \times 298)} = e^{13.29} = 5.9 \times 10^5$$

$K \gg 1$ Confirming the reaction strongly favours products at $298 \mathrm{ K}$.

Calculate $\Delta G$ for the reaction $\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \to 2\mathrm{NH_3(g)}$ at $298 \mathrm{ K}$ when $P(\mathrm{N_2)} = 10.0 \mathrm{ atm}$ $P(\mathrm{H_2)} = 30.0 \mathrm{ atm}$, $P(\mathrm{NH_3)} = 0.500 \mathrm{ atm}$.

$$Q = \frac{(0.500)^2}{(10.0)(30.0)^3} = \frac{0.250}{270000} = 9.26 \times 10^{-7}$$ $$\Delta G = \Delta G^\circ + RT\ln Q = -33000 + (8.314)(298)\ln(9.26 \times 10^{-7})$$ $$= -33000 + 2478 \times (-13.89) = -33000 - 34420 = -67420 \mathrm{ J/mol} = -67.4 \mathrm{ kJ/mol}$$

$\Delta G \lt 0$ So the reaction is spontaneous under these conditions. The high pressure of Reactants and low pressure of product drive the reaction forward.

1. Confusing $\Delta U$ and $\Delta H$. $\Delta H = \Delta U + P\Delta V$. They are equal only when there is no gas produced/consumed or when $\Delta V = 0$.
2. Wrong sign for work. In chemistry, $w = -P\Delta V$. When a gas expands ($\Delta V \gt 0$), the system does work on the surroundings ($w \lt 0$).
3. Forgetting that $\Delta H_f^\circ = 0$ for elements in their standard state. This is a convention.
4. Using the wrong sign convention for calorimetry. $q_{\mathrm{rxn}} = -q_{\mathrm{surroundings}}$.
5. Confusing entropy of the system with entropy of the universe. Spontaneity requires $\Delta S_{\mathrm{universe} \gt 0}$ Not just $\Delta S_{\mathrm{system} \gt 0}$.
6. Incorrect units in the Gibbs equation. $\Delta H$ is in kJ/mol; $\Delta S$ is in J/(mol$\cdot$K). Convert one of them before combining.
7. Using $R = 0.08206$ in $\Delta G^\circ = -RT\ln K$. Use $R = 8.314 \mathrm{ J/(mol\cdot\mathrm{K)}}$ because $\Delta G^\circ$ is in J/mol.
8. Assuming a negative $\Delta H$ guarantees spontaneity. If $\Delta S$ is sufficiently negative, $\Delta G$ can be positive even when $\Delta H$ is negative.
9. Forgetting that standard conditions are $298 \mathrm{ K}$ and $1 \mathrm{ atm}$ Not STP.

1. Calculate $\Delta H^\circ$ for $2\mathrm{Fe_2\mathrm{O_3(s)}} + 3\mathrm{C(s)} \to 4\mathrm{Fe(s)} + 3\mathrm{CO_2(g)}$ using standard enthalpies of formation.

2. When $3.50 \mathrm{ g}$ of $\mathrm{NaOH}$ is dissolved in $100.0 \mathrm{ g}$ of water in a calorimeter, the temperature rises from $23.0^\circ\mathrm{C}$ to $36.5^\circ\mathrm{C}$. Calculate $\Delta H$ per mole of $\mathrm{NaOH}$.

3. For a reaction with $\Delta H = 125 \mathrm{ kJ/mol}$ and $\Delta S = 200 \mathrm{ J/(mol\cdot\mathrm{K)}}$ Find the temperature range where the reaction is spontaneous.

4. Given $\Delta G_f^\circ$ values: $\mathrm{NO_2(g)} = 51.3 \mathrm{ kJ/mol}$ $\mathrm{N_2\mathrm{O_4(g)}} = 97.8 \mathrm{ kJ/mol}$. Find $\Delta G^\circ$ and $K$ for $2\mathrm{NO_2(g) \rightleftharpoons \mathrm{N_2\mathrm{O_4(g)}}}$ at $298 \mathrm{ K}$.

5. Estimate the enthalpy of combustion of $\mathrm{CH_4}$ using bond enthalpies. Compare with the value calculated from standard enthalpies of formation.

6. A reaction has $\Delta G^\circ = -20.0 \mathrm{ kJ/mol}$ at $298 \mathrm{ K}$. Calculate $K$.

7. Explain why the melting of ice is spontaneous above $0^\circ\mathrm{C}$ but not below, using $\Delta G = \Delta H - T\Delta S$.

8. Calculate $\Delta S^\circ$ for the reaction $2\mathrm{H_2(g)} + \mathrm{O_2(g)} \to 2\mathrm{H_2\mathrm{O(l)}}$ given: $S^\circ(\mathrm{H_2)} = 130.7$, $S^\circ(\mathrm{O_2)} = 205.1$ $S^\circ(\mathrm{H_2\mathrm{O, l)}} = 69.9 \mathrm{ J/(mol\cdot\mathrm{K)}}$.

9. A bomb calorimeter has $C_{\mathrm{cal}} = 850 \mathrm{ J/K}$. Burning $1.00 \mathrm{ g}$ of naphthalene ($\mathrm{C_{10}\mathrm{H_8}}$) raises the temperature by $2.46 \mathrm{ K}$. Calculate the enthalpy of combustion per mole of naphthalene.

10. For the reaction $\mathrm{NH_4\mathrm{NO_3(s)}} \to \mathrm{N_2\mathrm{O(g)}} + 2\mathrm{H_2\mathrm{O(g)}}$ $\Delta H^\circ = -36.0 \mathrm{ kJ/mol}$ and $\Delta S^\circ = 439 \mathrm{ J/(mol\cdot\mathrm{K)}}$. Is the reaction spontaneous at $298 \mathrm{ K}$? At what temperature does it become nonspontaneous?

11. Explain why the dissolution of $\mathrm{NH_4\mathrm{NO_3}}$ in water is endothermic yet spontaneous at room temperature.

12. Calculate $\Delta G$ (not $\Delta G^\circ$) for the reaction $\mathrm{N_2\mathrm{O_4(g) \rightleftharpoons 2\mathrm{NO_2(g)}}}$ at $298 \mathrm{ K}$ when $P(\mathrm{N_2\mathrm{O_4)}} = 0.50 \mathrm{ atm}$ and $P(\mathrm{NO_2)} = 0.10 \mathrm{ atm}$. Given: $\Delta G^\circ = 4.72 \mathrm{ kJ/mol}$.

13. Predict the sign of $\Delta S^\circ$ for each reaction and explain your reasoning: (a) $2\mathrm{Na(s)} + \mathrm{Cl_2(g)} \to 2\mathrm{NaCl(s)}$ (b) $\mathrm{CaCO_3(s)} \to \mathrm{CaO(s)} + \mathrm{CO_2(g)}$

14. Calculate $\Delta H^\circ$ for the reaction $\mathrm{N_2(g)} + 2\mathrm{O_2(g)} \to 2\mathrm{NO_2(g)}$ using the following data: $\frac{1}{2}\mathrm{N_2(g)} + \mathrm{O_2(g)} \to \mathrm{NO_2(g)}$ $\Delta H^\circ = 33.2 \mathrm{ kJ/mol}$.

15. A student calculates $\Delta G^\circ = -15 \mathrm{ kJ/mol}$ for a reaction at $298 \mathrm{ K}$ and concludes that the reaction will reach completion. Explain why this conclusion may not be justified.

16. Calculate $\Delta G^\circ$ at $500 \mathrm{ K}$ for a reaction with $\Delta H^\circ = -50 \mathrm{ kJ/mol}$ and $\Delta S^\circ = -80 \mathrm{ J/(mol\cdot\mathrm{K)}}$. Is the reaction spontaneous at this temperature?

17. Using the data below, calculate the standard enthalpy change for the reaction $\mathrm{C_2\mathrm{H_5\mathrm{OH(l)}}} + 3\mathrm{O_2(g)} \to 2\mathrm{CO_2(g)} + 3\mathrm{H_2\mathrm{O(l)}}$. $\Delta H_f^\circ(\mathrm{C_2\mathrm{H_5\mathrm{OH, l)}}} = -277.7 \mathrm{ kJ/mol}$.

18. Explain why a reaction with $\Delta H \gt 0$ and $\Delta S \gt 0$ is nonspontaneous at low temperatures but becomes spontaneous at high temperatures.

19. Calculate the boiling point of $\mathrm{Br_2}$ given that $\mathrm{Br_2(l)} \to \mathrm{Br_2(g)}$ has $\Delta H^\circ = 30.9 \mathrm{ kJ/mol}$ and $\Delta S^\circ = 93.2 \mathrm{ J/(mol\cdot\mathrm{K)}}$.

20. A calorimeter contains $200 \mathrm{ g}$ of water at $25.0^\circ\mathrm{C}$. When $5.00 \mathrm{ g}$ of $\mathrm{KOH}$ is dissolved, the temperature rises to $35.0^\circ\mathrm{C}$. Calculate the enthalpy of solution of $\mathrm{KOH}$ in kJ/mol.

21. For the reaction $2\mathrm{NO(g)} + \mathrm{O_2(g)} \to 2\mathrm{NO_2(g)}$ Given $\Delta H^\circ = -114.1 \mathrm{ kJ/mol}$ and $\Delta S^\circ = -146.5 \mathrm{ J/(mol\cdot\mathrm{K)}}$ Calculate the temperature above which the reaction is no longer spontaneous.

22. Using the following data, calculate $\Delta S^\circ$ for the reaction $4\mathrm{Fe(s)} + 3\mathrm{O_2(g)} \to 2\mathrm{Fe_2\mathrm{O_3(s)}}$: $S^\circ(\mathrm{Fe, s)} = 27.3$, $S^\circ(\mathrm{O_2, g)} = 205.1$ $S^\circ(\mathrm{Fe_2\mathrm{O_3, s)}} = 87.4 \mathrm{ J/(mol\cdot\mathrm{K)}}$.

23. Calculate the normal boiling point of chloroform ($\mathrm{CHCl_3}$) given that $\mathrm{CHCl_3(l)} \to \mathrm{CHCl_3(g)}$ has $\Delta H^\circ = 31.4 \mathrm{ kJ/mol}$ and $\Delta S^\circ = 94.2 \mathrm{ J/(mol\cdot\mathrm{K)}}$.

24. For the reaction $\mathrm{C(s)} + \mathrm{H_2\mathrm{O(g)}} \to \mathrm{CO(g)} + \mathrm{H_2(g)}$ $\Delta H^\circ = 131.3 \mathrm{ kJ/mol}$ and $\Delta S^\circ = 133.7 \mathrm{ J/(mol\cdot\mathrm{K)}}$. Calculate the minimum temperature at which this reaction becomes spontaneous.

25. Explain why the following statement is incorrect: “An exothermic reaction is always spontaneous.”

26. A bomb calorimeter with $C_{\mathrm{cal}} = 950 \mathrm{ J/K}$ is used to determine the enthalpy of combustion of benzoic acid ($\mathrm{C_7\mathrm{H_6\mathrm{O_2}}}$). Burning $1.00 \mathrm{ g}$ raises the temperature by $3.24 \mathrm{ K}$. Calculate the enthalpy of combustion per mole.

27. Calculate $\Delta G^\circ$ and $K$ at $298 \mathrm{ K}$ for the reaction $\mathrm{H_2(g)} + \mathrm{I_2(g)} \to 2\mathrm{HI(g)}$ given: $\Delta G_f^\circ(\mathrm{HI, g)} = 1.7 \mathrm{ kJ/mol}$.

28. For a certain reaction, $\Delta G^\circ = -5.4 \mathrm{ kJ/mol}$ at $300 \mathrm{ K}$. Calculate $K$ at this temperature and determine whether products or reactants are favoured.

29. Calculate the work done when $2.00 \mathrm{ mol}$ of a gas expands from $5.0 \mathrm{ L}$ to $15.0 \mathrm{ L}$ against a constant external pressure of $1.00 \mathrm{ atm}$.

30. Explain, using thermodynamic principles, why ice melts spontaneously at temperatures above $0^{\circ}\mathrm{C}$ even though the process is endothermic.

31. Calculate the work done by the system and $\Delta U$ when $3.00 \mathrm{ L}$ of gas at $2.00 \mathrm{ atm}$ expands against a constant external pressure of $0.50 \mathrm{ atm}$ to a final volume of $8.00 \mathrm{ L}$.

32. Given $\Delta H_f^\circ(\mathrm{NH_3, g)} = -46.1 \mathrm{ kJ/mol}$ $\Delta H_f^\circ(\mathrm{NO, g)} = 90.3 \mathrm{ kJ/mol}$ And $\Delta H_f^\circ(\mathrm{H_2\mathrm{O, g)}} = -241.8 \mathrm{ kJ/mol}$ Calculate $\Delta H^\circ$ for the reaction $4\mathrm{NH_3(g)} + 5\mathrm{O_2(g)} \to 4\mathrm{NO(g)} + 6\mathrm{H_2\mathrm{O(g)}}$.

33. The standard entropy values are: $S^\circ(\mathrm{C, s)} = 5.7$ $S^\circ(\mathrm{CO_2, g)} = 213.7$ $S^\circ(\mathrm{CO, g)} = 197.7 \mathrm{ J/(mol\cdot\mathrm{K)}}$. Calculate $\Delta S^\circ$ for the reaction $\mathrm{C(s)} + \mathrm{CO_2(g)} \to 2\mathrm{CO(g)}$ and comment on the sign.

Question 1: Hess's law and enthalpy of formation

Given the following data, calculate the standard enthalpy of formation of $\mathrm{CH_3\mathrm{OH(l)}}$:

• $\mathrm{C(s)} + \mathrm{O_2(g)} \to \mathrm{CO_2(g)}$, $\Delta H^\circ = -393.5 \mathrm{ kJ/mol}$
• $\mathrm{H_2(g)} + \frac{1}{2}\mathrm{O_2(g)} \to \mathrm{H_2\mathrm{O(l)}}$ $\Delta H^\circ = -285.8 \mathrm{ kJ/mol}$
• $\mathrm{CH_3\mathrm{OH(l)}} + \frac{3}{2}\mathrm{O_2(g)} \to \mathrm{CO_2(g)} + 2\mathrm{H_2\mathrm{O(l)}}$ $\Delta H^\circ = -726.4 \mathrm{ kJ/mol}$

Answer

Target: $\mathrm{C(s)} + 2\mathrm{H_2(g)} + \frac{1}{2}\mathrm{O_2(g)} \to \mathrm{CH_3\mathrm{OH(l)}}$

Manipulate the given equations:

(1) $\mathrm{C(s)} + \mathrm{O_2(g)} \to \mathrm{CO_2(g)}$ $\Delta H^\circ = -393.5 \mathrm{ kJ/mol}$ — keep

(2) $2\mathrm{H_2(g)} + \mathrm{O_2(g)} \to 2\mathrm{H_2\mathrm{O(l)}}$ $\Delta H^\circ = 2(-285.8) = -571.6 \mathrm{ kJ/mol}$ — multiply by 2

(3) $\mathrm{CO_2(g)} + 2\mathrm{H_2\mathrm{O(l)}} \to \mathrm{CH_3\mathrm{OH(l)}} + \frac{3}{2}\mathrm{O_2(g)}$ $\Delta H^\circ = +726.4 \mathrm{ kJ/mol}$ — reverse

Add (1) + (2) + (3):

$\mathrm{C(s)} + 2\mathrm{H_2(g)} + \frac{1}{2}\mathrm{O_2(g)} \to \mathrm{CH_3\mathrm{OH(l)}}$

$\Delta H_f^\circ = -393.5 + (-571.6) + 726.4 = -238.7 \mathrm{ kJ/mol}$.

Question 2: Gibbs free energy and spontaneity

For the reaction $\mathrm{NH_4\mathrm{NO_3(s)}} \to \mathrm{N_2\mathrm{O(g)}} + 2\mathrm{H_2\mathrm{O(g)}}$ $\Delta H^\circ = -36.0 \mathrm{ kJ/mol}$ and $\Delta S^\circ = 347 \mathrm{ J/(mol\cdot K)}$. Calculate $\Delta G^\circ$ at $298 \mathrm{ K}$ and determine the temperature range over which the reaction is spontaneous.

Answer

$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -36,000 - 298 \times 347 = -36,000 - 103,406 = -139,406 \mathrm{ J/mol} = -139.4 \mathrm{ kJ/mol}$.

Since $\Delta G^\circ \lt 0$ at $298 \mathrm{ K}$ The reaction is spontaneous at this temperature.

The reaction is spontaneous when $\Delta G \lt 0$:

$\Delta H - T\Delta S \lt 0$

Since $\Delta H = -36.0 \mathrm{ kJ/mol}$ (negative) and $\Delta S = +347 \mathrm{ J/(mol\cdot K)}$ (positive), both terms favour spontaneity. The reaction is spontaneous at all temperatures. There is no upper temperature limit because the $-T\Delta S$ term always contributes negatively to $\Delta G$ when $\Delta S$ is positive.

Question 3: Calorimetry and specific heat

A $50.0 \mathrm{ g}$ sample of an unknown metal is heated to $100.0^\circ\mathrm{C}$ and then placed in $100.0 \mathrm{ g}$ of water at $25.0^\circ\mathrm{C}$ in a coffee-cup calorimeter. The final Temperature of the mixture is $28.8^\circ\mathrm{C}$. Calculate the specific heat capacity of the metal. Assume no heat loss to the calorimeter.

Answer

Heat gained by water = heat lost by metal.

$q_{\mathrm{water}} = m_{\mathrm{water} \times c_{\mathrm{water} \times \Delta T_{\mathrm{water}}}} = 100.0 \times 4.184 \times (28.8 - 25.0) = 100.0 \times 4.184 \times 3.8 = 1589.9 \mathrm{ J}$.

$q_{\mathrm{metal}} = m_{\mathrm{metal} \times c_{\mathrm{metal} \times \Delta T_{\mathrm{metal}}}} = 50.0 \times c_{\mathrm{metal} \times (100.0} - 28.8) = 50.0 \times c_{\mathrm{metal} \times 71.2}$.

Setting equal: $1589.9 = 50.0 \times c_{\mathrm{metal} \times 71.2}$.

$c_{\mathrm{metal}} = 1589.9 / 3560 = 0.447 \mathrm{ J/(g\cdot^\circ\mathrm{C)}}$.

This value is close to that of iron ($0.449 \mathrm{ J/(g\cdot^\circ\mathrm{C)}}$), suggesting the unknown metal may be iron.

Question 4: Entropy change of surroundings

For the vaporisation of water at $100^\circ\mathrm{C}$ and $1 \mathrm{ atm}$: $\mathrm{H_2\mathrm{O(l)}} \to \mathrm{H_2\mathrm{O(g)}}$ $\Delta H_{\mathrm{vap}} = 40.7 \mathrm{ kJ/mol}$. Calculate $\Delta S_{\mathrm{system}}$ $\Delta S_{\mathrm{surroundings}}$ And $\Delta S_{\mathrm{universe}}$. Is the process spontaneous At this temperature?

Answer

$\Delta S_{\mathrm{system}} = \Delta H_{\mathrm{vap} / T} = 40,700 / 373 = 109.1 \mathrm{ J/(mol\cdot K)}$.

$\Delta S_{\mathrm{surroundings}} = -\Delta H_{\mathrm{vap} / T} = -40,700 / 373 = -109.1 \mathrm{ J/(mol\cdot K)}$.

$\Delta S_{\mathrm{universe}} = \Delta S_{\mathrm{system}} + \Delta S_{\mathrm{surroundings}} = 109.1 + (-109.1) = 0 \mathrm{ J/(mol\cdot K)}$.

At $100^\circ\mathrm{C}$ and $1 \mathrm{ atm}$ Liquid and gaseous water are in equilibrium, so $\Delta G = 0$ and $\Delta S_{\mathrm{universe}} = 0$. The process is at equilibrium, not Spontaneous in either direction. Above $100^\circ\mathrm{C}$ Vaporisation becomes spontaneous ($\Delta S_{\mathrm{universe} \gt 0}$).

Question 5: Bond enthalpy calculation

Using the following average bond enthalpies, estimate $\Delta H$ for the reaction $\mathrm{CH_4(g)} + 2\mathrm{Cl_2(g)} \to \mathrm{CH_2\mathrm{Cl_2(g)}} + 2\mathrm{HCl(g)}$:

C-H: $413 \mathrm{ kJ/mol}$ Cl-Cl: $242 \mathrm{ kJ/mol}$ C-Cl: $339 \mathrm{ kJ/mol}$ H-Cl: $431 \mathrm{ kJ/mol}$.

Answer

Bonds broken (reactants):

• 4 C-H bonds in $\mathrm{CH_4}$: $4 \times 413 = 1652 \mathrm{ kJ/mol}$
• 2 Cl-Cl bonds: $2 \times 242 = 484 \mathrm{ kJ/mol}$

Wait — not all C-H bonds break. In $\mathrm{CH_2\mathrm{Cl_2}}$ Two C-H bonds remain. So only 2 C-H bonds break.

Corrected bonds broken:

• 2 C-H bonds: $2 \times 413 = 826 \mathrm{ kJ/mol}$
• 2 Cl-Cl bonds: $2 \times 242 = 484 \mathrm{ kJ/mol}$
• Total broken: $826 + 484 = 1310 \mathrm{ kJ/mol}$

Bonds formed (products):

• 2 C-Cl bonds: $2 \times 339 = 678 \mathrm{ kJ/mol}$
• 2 H-Cl bonds: $2 \times 431 = 862 \mathrm{ kJ/mol}$
• Total formed: $678 + 862 = 1540 \mathrm{ kJ/mol}$

$\Delta H = \mathrm{bonds broken} - \mathrm{bonds formed} = 1310 - 1540 = -230 \mathrm{ kJ/mol}$.

The reaction is exothermic because stronger bonds (H-Cl, C-Cl) are formed than are broken (C-H, Cl-Cl).

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: $mgh = \frac{1}{2}mv^2$

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

Physics explores the fundamental rules governing matter, energy, space, and time. At its heart lies the principle that complex phenomena emerge from simple interactions - gravity shapes orbits, electromagnetism binds atoms, and quantum mechanics governs the subatomic realm. Understanding these laws allows us to build technologies from smartphones to spacecraft and to comprehend our place in the cosmos.

• Mechanics
• Waves
• Electricity
• Fields
• AP Physics, Work, Energy, and Power: The work-energy theorem from physics underpins enthalpy and internal energy calculations in thermodynamics.