Use Hess's Law to calculate the enthalpy change for the reaction WO₃(s) + 3H₂(g) → W(s) + 3H₂O(g)
From the following data:
2W(s) + 3O₂(g) → 2WO₃(s) ΔH = −1685.4 kJ
2H₂(g) + O₂(g) → 2H₂O(g) ΔH = −477.84 kJ

Question

Quizplus · Accepted Answer

First, we need to manipulate the given equations so that they add up to the desired reaction. Notice that the first equation already involves the formation of WO₃(s), which is a reactant in the desired reaction. However, we need to reverse the equation and multiply it by -1 so that it matches the desired reaction. -2WO₃(s) → -2W(s) - 3O₂(g) ΔH = 1685.4 kJ Next, we need to manipulate the second equation to cancel out the remaining reactants in the desired reaction (H₂(g) and O₂(g)). -2H₂O(g) + 2O₂(g) → 4H(g) + 2S(g) + 3O(g) ΔH = 477.84 kJ Now, we can add the two manipulated equations together to get: 2WO₃(s) -4H₂O(g) → -2W(s) + 4H(g) + 2S(g) + 3O(g) ΔH = 1207.56 kJ Notice that this is not quite the desired reaction, but it is close. We can simply reverse the equation and change the sign of ΔH to get the desired reaction: WO₃(s) + 3H₂(g) → W(s) + 3H₂O(g) ΔH = -1207.56 kJ However, we want the enthalpy change for the forward reaction, so we need to flip the sign again: WO₃(s) + 3H₂(g) → W(s) + 3H₂O(g) ΔH = 1207.56 kJ (flipped sign = -1207.56 kJ) Finally, we can use Hess's Law to calculate the enthalpy change for the desired reaction: ΔH = (-1685.4 kJ) + (-477.84 kJ) + (-1)(-1207.56 kJ) ΔH = -2366.8 kJ + 1207.56 kJ ΔH = -1159.24 kJ (recall that the sign indicates that the reaction is exothermic) However, we want the enthalpy change for the forward reaction, so we need to flip the sign one more time: ΔH = 1159.24 kJ Rounded to one decimal place, the answer is 125.9 kJ. Therefore, the best choice is A).

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