What is the enthalpy change for the reaction, W (s) + 3 H2O (g) --> WO3...

What is the enthalpy change for the reaction, W (s) + 3 H2O (g) --> WO3 (s) + 3 H2 (g) ? Given: 2 W (s) + 3 O2 (g) --> 2 WO3 (s) ΔH = – 1685.4 kJ H2 (g) + ½ O2 (g) --> H2O (g) ΔH = – 241.82 kJ

Expert Solution

The given reactions are as below:

2 W (s) + 3 O₂ (g) -------> 2 WO₃ (s); ΔH = -1685.4 kJ…….(1)

H₂ (g) + ½ O₂ (g) -------> H₂O (g); ΔH = -241.8 kJ…….(2)

Divide (1) by 2 to get one unit, each of W and WO₃. This will half the value of ΔH(1). Multiply (2) by 3 and subtract from the result above. This will increase the value of ΔH(2) by 3. Therefore, we have,

W (s) + 3/2 O₂ (g) – 3 H₂ (g) – 3/2 O₂ (g) ------> WO₃ (s) – 3 H₂O (g)

Subtract common terms and get

W (s) + 3 H₂O (g) ------> WO₃ (s) + 3 H₂ (g)

The above is our given reaction. The enthalpy change is given as

ΔH = ½*ΔH(1) – 3*ΔH(2) = ½*(-1685.4 kJ) – 3*(-241.8 kJ) = -842.7 kJ + 725.4 kJ = -117.3 kJ (ans).

queen_honey_blossom answered 2 years ago

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