Question

Определить стандартную энтальпию образования этилена (в килоджоулях), используя следующие данные:С2Н4(г) + 3О2(г) = 2СО2(г) + 2Н2О(р); ΔН0= –1323 кДж; С(графіт) + О2(г) = СО2(г); ΔН0= –393,5 кДж;Н2(г) + 1/2О2(г) = Н2О(р); ΔН0= –241,8 кДж Ответ дать с точностью до десятых.

Answer 1

Ух ты! Это ж мои любимые задачки)))

Answer 2

Standard Enthalpy of Formation of Ethylene

To determine the standard enthalpy of formation of ethylene (C2H4) in kilojoules, we can use the given chemical equation and the standard enthalpies of formation of the reactants and products involved.

The given chemical equation is: C2H4(g) + 3O2(g) = 2CO2(g) + 2H2O(l)

The standard enthalpy change (ΔH°) for this reaction is -1323 kJ.

To calculate the standard enthalpy of formation of ethylene, we need to consider the standard enthalpies of formation of the reactants and products involved.

The standard enthalpy of formation (ΔH°f) is the enthalpy change when one mole of a compound is formed from its elements in their standard states, with all substances in their standard states at 25°C and 1 atm pressure.

Using the given equation, we can break down the reaction into the formation of individual compounds:

1. Formation of carbon dioxide (CO2): C(graphite) + O2(g) = CO2(g) The standard enthalpy change for this reaction is -393.5 kJ/mol [[1]].

2. Formation of water (H2O): H2(g) + 1/2O2(g) = H2O(l) The standard enthalpy change for this reaction is -241.8 kJ/mol [[2]].

Now, we can calculate the standard enthalpy of formation of ethylene (C2H4) using the given equation and the standard enthalpies of formation of the reactants and products:

ΔH°f(C2H4) = [2 × ΔH°f(CO2)] + [2 × ΔH°f(H2O)] - [ΔH°]

Substituting the values:

ΔH°f(C2H4) = [2 × (-393.5 kJ/mol)] + [2 × (-241.8 kJ/mol)] - (-1323 kJ) ΔH°f(C2H4) = -787 kJ/mol + (-483.6 kJ/mol) + 1323 kJ ΔH°f(C2H4) = 52.4 kJ/mol

Therefore, the standard enthalpy of formation of ethylene (C2H4) is 52.4 kJ/mol.

Please note that the answer is rounded to one decimal place as requested.

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