Calculate the enthalpy change for the reaction C2H4 + H2 -> C2H6 using the provided bond energies.

Here are the solutions for the remaining parts of the question: ii) A: Use this information to calculate the enthalpy change for the process The reaction is H_2C=CH_2 + H-H → H_3C-CH_3. The bond energies provided are: C-C = 150 kJ/mol C=C = 610 kJ/mol H-H = 436 kJ/mol C-H = 410 kJ/mol Step 1: Calculate the energy required to break bonds (reactants). Bonds broken: • 1 × C=C bond in C_2H_4 = 1 × 610 kJ/mol = 610 kJ/mol • 4 × C-H bonds in C_2H_4 = 4 × 410 kJ/mol = 1640 kJ/mol • 1 × H-H bond in H_2 = 1 × 436 kJ/mol = 436 kJ/mol Total energy to break bonds = 610 + 1640 + 436 = 2686 kJ/mol Step 2: Calculate the energy released when bonds are formed (products). Bonds formed: • 1 × C-C bond in C_2H_6 = 1 × 150 kJ/mol = 150 kJ/mol • 6 × C-H bonds in C_2H_6 = 6 × 410 kJ/mol = 2460 kJ/mol Total energy released when bonds form = 150 + 2460 = 2610 kJ/mol Step 3: Calculate the enthalpy change ( H). H = Energy broken - Energy formed H = 2686 kJ/mol - 2610 kJ/mol = +76 kJ/mol B: The enthalpy change of this reaction was found by experiment to be -136 kJ/mol. Explain why this value is different from that calculated above. The calculated enthalpy change uses average bond energies, which are values averaged over many different compounds. The actual bond energies in specific molecules like ethene and ethane can differ from these average values. Additionally, bond energies are typically defined for substances in the gaseous state, and the experimental conditions might not perfectly match this. d) Complete the table below by filling in either the crystal type or example. | Crystal type | Example | | :----------------------- | :------------- | | Body-centred cubic | Iron (Fe) | | Hexagonal close packed | Magnesium (Mg) | | Giant covalent structure | Diamond | e) i) State Le Chatelier's Principle. Le Chatelier's Principle states that if a change of condition (such as temperature, pressure, or concentration) is applied to a system in equilibrium, the system will shift in a direction that counteracts or relieves the stress, re-establishing a new equilibrium. ii) Consider the reaction 2NO_2(g) N_2O_4(g) H^ = +58.1 kJ/mol How does an increase in temperature affect: A. The equilibrium constant? The reaction is endothermic ( H^ = +58.1 kJ/mol). An increase in temperature favors the endothermic direction (forward reaction) to absorb the added heat. This leads to a higher concentration of products and a lower concentration of reactants at equilibrium. Therefore, the equilibrium constant will increase. B. The equilibrium position? An increase in temperature will shift the equilibrium position to the right, favoring the formation of N_2O_4. That's 2 down. 3 left today — send the next one.