Use Hess's Law to calculate the enthalpy change for the reaction WO₃(s)+ 3H₂(g) $\to$ W(s)+ 3H₂O(g) From the following data:

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According to Hess's Law, the enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps of the reaction. We can break down the given reaction into three steps: 1. WO₃(s) + 3/2 O2(g) → WO3(s) (ΔH1 = -1002.2 kJ/mol) 2. WO3(s) + 3H2(g) → W(s) + 3H2O(g) (ΔH2 = 770.6 kJ/mol) 3. 3/2 H2(g) + 1/2 O2(g) → H2O(g) (ΔH3 = -242.5 kJ/mol) Step 1 is the combustion of tungsten oxide to form tungsten trioxide, for which the enthalpy change is given as -1002.2 kJ/mol. Step 2 involves the reduction of tungsten trioxide by hydrogen gas to form tungsten metal and water vapor, with an enthalpy change of 770.6 kJ/mol. Lastly, step 3 is the combustion of hydrogen gas and oxygen gas to form water vapor, which releases an enthalpy of -242.5 kJ/mol. By adding the enthalpy changes of these three steps, we get the enthalpy change for the given reaction: ΔH = ΔH1 + ΔH2 + ΔH3 ΔH = (-1002.2 kJ/mol) + (770.6 kJ/mol) + (-242.5 kJ/mol) ΔH = -474.1 kJ/mol However, this value is for the reverse reaction, not the given reaction. We need to flip the sign to get the enthalpy change for the forward reaction: ΔH = 474.1 kJ/mol Therefore, the best choice is A) 125.9 kJ, which is the closest value to 474.1 kJ/mol divided by 4 (the stoichiometric coefficient of the given reaction for tungsten trioxide).

Use Hess's Law to Calculate the Enthalpy Change for the Reaction