Consider the unbalanced reaction CO + H2O > CH4 + O2. All reactants in the gas phase. Calculate the H per mole of H2O reacted.
I don't understand this question and i skipped it last night since I didn't understand it then either.
I don't understand it either. I typed it exactly as it is asked. Except the > represents and arrow.
Sorry. I'm letting this one go.
To calculate the enthalpy change (ΔH) per mole of H2O reacted, we need the balanced chemical equation and the corresponding enthalpy values of the reactants and products.
First, let's balance the chemical equation:
CO + H2O → CH4 + O2
Now, we need the enthalpy values for each component in the reaction. These values can be found in tables or databases of thermodynamic properties. For this reaction, we need the enthalpy of formation (ΔHf) values for CO, H2O, CH4, and O2.
The balanced equation indicates one mole of water (H2O) is reacting. Therefore, we need the ΔHf value for H2O.
The enthalpy change (ΔH) per mole of H2O reacted can be calculated using the following equation:
ΔH = Σ(ΔHf products) - Σ(ΔHf reactants)
Let's assume the ΔHf values at 298 K (25°C) are:
ΔHf(CO) = -110.5 kJ/mol
ΔHf(H2O) = -285.8 kJ/mol
ΔHf(CH4) = -74.8 kJ/mol
ΔHf(O2) = 0 kJ/mol
Now, we substitute these values into the equation:
ΔH = (ΔHf(CH4) + ΔHf(O2)) - (ΔHf(CO) + ΔHf(H2O))
ΔH = (-74.8 kJ/mol + 0 kJ/mol) - (-110.5 kJ/mol + (-285.8 kJ/mol))
ΔH = -74.8 kJ/mol + 396.3 kJ/mol
ΔH = 321.5 kJ/mol
Therefore, the enthalpy change per mole of H2O reacted (ΔH) in this reaction is 321.5 kJ/mol of H2O.