Question

C2H4+3O2=2CO2+2H2O+1400 кДж n(O2) 1 моль н.у. Q-?

Answer 1

Відповідь:

3моль  ---14000

1моль------х

х=1400/3=466,7кДж

Пояснення:

Answer 2

To calculate the heat released (Q) during the combustion of 1 mole of ethylene (C2H4) with 3 moles of oxygen (O2) at standard conditions (n.u.), we need to find the balanced chemical equation and then use the appropriate enthalpy values for the reactants and products involved.

The balanced chemical equation for the combustion of ethylene (C2H4) with oxygen (O2) is:

C2H4 + 3 O2 → 2 CO2 + 2 H2O

First, let's calculate the change in enthalpy (ΔH) for the reaction:

ΔH = Σ(ΔH_products) - Σ(ΔH_reactants)

The enthalpy change for each compound can be found in standard enthalpy of formation tables. The values for standard enthalpies of formation are typically given in kJ/mol.

ΔH_f° (C2H4) = +52.3 kJ/mol (standard enthalpy of formation of ethylene) ΔH_f° (O2) = 0 kJ/mol (standard enthalpy of formation of oxygen, since it is in its elemental form) ΔH_f° (CO2) = -393.5 kJ/mol (standard enthalpy of formation of carbon dioxide) ΔH_f° (H2O) = -285.8 kJ/mol (standard enthalpy of formation of water)

Now, we can plug these values into the equation:

ΔH = [2 * ΔH_f° (CO2) + 2 * ΔH_f° (H2O)] - [ΔH_f° (C2H4) + 3 * ΔH_f° (O2)] ΔH = [2 * (-393.5 kJ/mol) + 2 * (-285.8 kJ/mol)] - [52.3 kJ/mol + 3 * 0 kJ/mol] ΔH = [-787 kJ/mol - 571.6 kJ/mol] - 52.3 kJ/mol ΔH = -1358.6 kJ/mol - 52.3 kJ/mol ΔH = -1410.9 kJ/mol

The negative sign indicates that the reaction is exothermic, meaning it releases heat. The magnitude of ΔH is 1410.9 kJ/mol.

Since the reaction involves 1 mole of O2, the heat released (Q) for the combustion of 1 mole of ethylene with 1 mole of O2 at standard conditions is 1410.9 kJ.

Please note that the 1400 kJ value provided in the initial question might not be the correct enthalpy value for this specific reaction. The correct value is calculated as shown above.