### Related Formula The enthalpy of formation reaction can be evaluated by balancing atomization and bond cleavage details: Delta H_f^circ = sum Delta Htextatomization (reactants) - sum Delta Htextbonds broken/formed (products) $\Delta H_f^{\circ} = \sum \Delta H{\text{atomization (reactants)}} - \sum \Delta H{\text{bonds broken/formed (products)}} $ ### Core Logic The target formation reaction for ethylene (textC_2textH_4$\text{C}_2\text{H}_4$) from standard elemental states is: 2C(textgraphite) + 2H_2(g) ightarrow CH_2=CH_2(g) $ 2C(\text{graphite}) + 2H_2(g) ightarrow CH_2=CH_2(g) $ To construct this pathway: 1. Sublime 2 moles of solid graphite: 2 times Delta H_textsub^circ[C]$2 \times \Delta H_{\text{sub}}^{\circ}[C]$ 2. Dissociate 2 moles of gaseous textH-textH$\text{H}-\text{H}$ bonds: 2 times Delta_H-HH^circ$2 \times \Delta_{H-H}H^{\circ}$ 3. Form 1 mole of textC=textC$\text{C}=\text{C}$ double bonds: -1 times Delta_C=CH^circ$-1 \times \Delta_{C=C}H^{\circ}$ 4. Form 4 moles of textC-textH$\text{C}-\text{H}$ single bonds: -4 times Delta_C-HH^circ$-4 \times \Delta_{C-H}H^{\circ}$ ### Step 1: Arithmetic Calculation Delta H_f^circ = [2 times 710] + [2 times 436] - 611 - [4 times 414] $ Delta H_f^{\circ} = [2 \times 710] + [2 \times 436] - 611 - [4 \times 414] $ Delta H_f^circ = 1420 + 872 - 611 - 1656 = 2292 - 2267 = 25text kJ mol^-1 $ Delta H_f^{\circ} = 1420 + 872 - 611 - 1656 = 2292 - 2267 = 25\text{ kJ mol}^{-1} $ ### Pattern Recognition Shortcut: Group energy components systematically. Reactant state atomization costs +2292text kJ$+2292\text{ kJ}$. Exothermic structural bond formation releases -2267text kJ$-2267\text{ kJ}$. Net difference yields a small endothermic value of +25text kJ/mol$+25\text{ kJ/mol}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Thermodynamics