Given the following data:2W (s) + 3O2 (g) → 2WO3 (s) ΔHrxn = −1685.4 kJ/mol rxn2H2 (g) + O2 (g) → 2H2O (g) ΔHrxn = −477.84 kJ/mol rxnDetermine the enthalpy change for the reaction WO3 (s) + 3H2 (g) → W (s) + 3H2O (g)
Question
Given the following data:2W (s) + 3O2 (g) → 2WO3 (s) ΔHrxn = −1685.4 kJ/mol rxn2H2 (g) + O2 (g) → 2H2O (g) ΔHrxn = −477.84 kJ/mol rxnDetermine the enthalpy change for the reaction WO3 (s) + 3H2 (g) → W (s) + 3H2O (g)
Solution
To determine the enthalpy change for the reaction WO3 (s) + 3H2 (g) → W (s) + 3H2O (g), we need to manipulate the given reactions and their enthalpies to match the desired reaction.
The given reactions are:
- 2W (s) + 3O2 (g) → 2WO3 (s) ΔHrxn = −1685.4 kJ/mol rxn
- 2H2 (g) + O2 (g) → 2H2O (g) ΔHrxn = −477.84 kJ/mol rxn
The desired reaction is:
- WO3 (s) + 3H2 (g) → W (s) + 3H2O (g)
First, we need to reverse the first reaction and divide it by 2 to get WO3 (s) → W (s) + 3/2 O2 (g). The enthalpy change of this reaction is half of the original reaction and the sign is reversed, so ΔH = 1685.4/2 = 842.7 kJ/mol.
Next, we need to multiply the second reaction by 3/2 to get 3H2 (g) + 3/2 O2 (g) → 3H2O (g). The enthalpy change of this reaction is 1.5 times the original reaction, so ΔH = 1.5 * -477.84 = -716.76 kJ/mol.
Finally, we add these two reactions to get the desired reaction:
WO3 (s) + 3H2 (g) → W (s) + 3H2O (g)
The enthalpy change for the desired reaction is the sum of the enthalpy changes of the two manipulated reactions, so ΔH = 842.7 kJ/mol - 716.76 kJ/mol = 125.94 kJ/mol.
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