The given balanced chemical equation is:
[C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)]
We need to write the expression for the equilibrium constant ((K_{\text{eq}})) for this reaction
How much energy is released when 23.0 of ethylene reacts? Be sure your answer has the correct number of significant figures.
Given the balanced chemical equation:
C2H4(g) + H2(g) → C2H6(g)
The equilibrium constant expression (Keq) can be written as:
Keq = [C2H6] / ([C2H4] * [H2])
We can assume that initially there is only 23.0g of ethylene (C2H4) present in the system. We also assume that all the ethylene is consumed during the reaction.
First, we need to convert grams of ethylene to moles:
Molar mass of C2H4 = 2*12.01 + 4*1.01 = 28.05 g/mol
Number of moles of C2H4 = 23.0g / 28.05 g/mol = 0.820 mol
Since the stoichiometry of the reaction is 1:1 for C2H4 and C2H6, the number of moles of C2H6 formed will also be 0.820 mol.
Now, we need to calculate the energy released during the reaction. The enthalpy change (ΔH) for the reaction can be found in a thermochemical table or calculated using bond energies.
Assuming that the reaction is exothermic, the energy released can be calculated using the following formula:
Energy released = n * ΔH
Given that 1 mol of C2H6 releases 136.0 kJ of energy, we can calculate the energy released when 0.820 mol of C2H6 is formed:
Energy released = 0.820 mol * 136.0 kJ/mol = 111.52 kJ
Therefore, when 23.0g of ethylene reacts, 111.52 kJ of energy is released. The answer has the correct number of significant figures.