Topic 6: Energetics

Standard conditions use a pressure of 100 kPa and a specified temperature, usually 298 K. Every substance must be in its standard state at those conditions unless another state is stated.

Physical state matters because changing state also transfers energy. For example, combustion data for liquid pentane include a different starting enthalpy from data for gaseous pentane.

The sign refers to the reacting system. If an insulated solution warms during a reaction, the solution gains heat but the reaction system releases it, so the reaction $\Delta H$ is negative.

A value of $\Delta H=-200$ kJ mol$^{-1}$ means 200 kJ is released per mole of reaction as written. A value of +50 kJ mol$^{-1}$ means 50 kJ is absorbed.

Label the vertical axis enthalpy, write the correct species and states on horizontal reactant and product levels, and draw the $\Delta H$ arrow directly between those levels with its value and units.

The vertical separation represents the enthalpy change. For SO$_3$(g)+H$_2$O(l)$\rightarrow$H$_2$SO$_4$(aq), $\Delta H=-200$ kJ mol$^{-1}$, reactants must be 200 kJ mol$^{-1}$ above products.

Definitions determine equation coefficients. For formation of one mole of water(l): H$_2$(g)+$\tfrac12$O$_2$(g)$\rightarrow$H$_2$O(l). For atomisation of chlorine: $\tfrac12$Cl$_2$(g)$\rightarrow$Cl(g).

Before choosing data, check: correct amount (usually 1 mol of the named product or substance), complete combustion where required, elemental standard states for formation/atomisation, and every physical state.

For the material whose temperature is measured, $q=mc\Delta T$, where $q$ is in J, $m$ in g, $c$ in J g$^{-1}$ °C$^{-1}$ and $\Delta T$ in °C.

If 51.0 g of solution warms by 6.9 °C and $c=4.18$ J g$^{-1}$ °C$^{-1}$, $q=51.0\times4.18\times6.9=1.47\times10^3$ J. If 0.0273 mol reacted, $\Delta H=-1.47/(0.0273)=-53.9$ kJ mol$^{-1}$.

Write the target equation, arrange known equations so unwanted species cancel, reverse any equation that runs the wrong way and change the sign of its $\Delta H$, multiply equations and enthalpies by the same factor, then add.

Apply stoichiometric coefficients to every enthalpy value and preserve physical states. Confirm that adding the manipulated chemical equations gives exactly the target before adding their numbers.

For CaCO$_3$(s)$\rightarrow$CaO(s)+CO$_2$(g), measure $\Delta H_1$ for CaCO$_3$+2HCl and $\Delta H_2$ for CaO+2HCl to their common CaCl$_2$(aq)+H$_2$O(l) destination. Then $\Delta H_{target}=\Delta H_1-\Delta H_2$.

Use the same acid concentration and comparable total solution mass, ensure the chosen reagent is fully reacted, stir consistently, and repeat measurements. Record enough temperature-time data for a cooling correction.

Record temperature at fixed times before mixing, add reactants at a known time, continue readings after the maximum, plot temperature against time, fit the post-reaction cooling line and extrapolate it back to the mixing time. Use the extrapolated temperature to obtain corrected $\Delta T$.

State the mechanism of each error, its direction where defensible, and whether it is random or systematic. Compare repeats and calculate percentage uncertainty from apparatus uncertainties rather than calling every difference 'human error'.

Use $\Delta H\approx\sum E(bonds\ broken)-\sum E(bonds\ formed)$. Breaking bonds requires energy and contributes positively; forming bonds releases energy and is subtracted.

Draw complete structures, count each bond in the stoichiometric equation, multiply by its mean value, total reactant bonds broken and product bonds formed, then subtract with units kJ mol$^{-1}$.

Mean values average different molecular environments and refer to gaseous species. They therefore give an estimate; state changes and the actual bond environment can make the value differ from an experimental standard enthalpy.

Start with a balanced reaction and $\Delta H=\sum E_{broken}-\sum E_{formed}$. Count the unknown bond as $nX$, where $n$ is the total number of those bonds formed or broken in the stoichiometric reaction.

If the products contain 48 S–F bonds in total, keep their contribution as $48X$ until all known bond totals and the reaction $\Delta H$ have been inserted, then divide by 48 to obtain the mean S–F bond enthalpy.

The result should be a positive energy per mole of bonds. Recount bonds, coefficients and whether the unknown is on the broken or formed side if the sign or size is implausible.

A smaller bond enthalpy means less energy is required for homolytic bond breaking, so that bond may break more readily and may help a reaction proceed faster at room temperature. A larger value indicates a stronger bond that is harder to break.

Rate depends on the complete mechanism, activation energy, collision geometry, temperature and catalysts. Bond enthalpies are averaged gas-phase data and indicate only one energetic contribution.