Respuesta :
Answer:
Explanation:
To balance the given thermochemical equation:
CH2O2(g) + H2(g) ⟶ H2O(g) + CH4(g)
The balanced equation is:
CH2O2(g) + 4H2(g) ⟶ 2H2O(g) + CH4(g)
Now, let's determine the enthalpy of reaction, ΔrH∘, using the bond enthalpies table.
The enthalpy change for a reaction can be calculated using the bond enthalpies of the bonds broken and formed in the reaction. The bond enthalpy is the energy required to break a specific bond.
Given bond enthalpies:
C-H = 413 kJ/mol
O-H = 463 kJ/mol
C=O = 799 kJ/mol
H-H = 436 kJ/mol
Calculate the total bond energy for the bonds broken and formed:
Bonds broken:
1 mol of C-H bonds (4 x 413 kJ/mol) + 1 mol of H-H bonds (1 x 436 kJ/mol)
Bonds formed:
2 mol of O-H bonds (2 x 463 kJ/mol) + 1 mol of C-H bonds (1 x 413 kJ/mol)
Calculate the total energy change:
ΔrH∘ = Σ(bonds broken) - Σ(bonds formed)
ΔrH∘ = (4 x 413 + 1 x 436) - (2 x 463 + 1 x 413)
ΔrH∘ = (1652 + 436) - (926 + 413)
ΔrH∘ = 2088 - 1339
ΔrH∘ = 749 kJ/mol
Therefore, the enthalpy of reaction, ΔrH∘, is 749 kJ/mol.
To calculate the heat exchanged when 37.76 g of CH2O2(g) reacts, you need to first determine the limiting reactant.
Calculate the moles of CH2O2:
n = m/M
n = 37.76 g / 46 g/mol
n = 0.8217 mol
Using the balanced equation:
1 mol of CH2O2 reacts with 4 mol of H2.
So, 0.8217 mol of CH2O2 would react with 0.8217 mol x 4 = 3.2868 mol of H2.
Now, calculate the heat exchanged using the enthalpy of reaction:
q = n x ΔrH∘
q = 3.2868 mol x 749 kJ/mol
q = 2466.09 kJ
Therefore, the heat exchanged when 37.76 g of CH2O2(g) reacts is 2466.09 kJ.
