Question

using hesss law, what is δh° rxn for the following reaction?
wo₃(s) + 3h₂(g) → w(s) + 3h₂o(g)
2w(s) + 3o₂(g) → 2wo₃(s) δh° rxn = -1685.8 kj/mol
2h₂(g) + o₂(g) → 2h₂o(g) δh° rxn = -483.6 kj/mol
o117.5 kj/mol
o235.0 kj/mol
o359.3 kj/mol
o1202.2 kj/mol
o2169.4 kj/mol

Explanation:

Step1: Identify the target reaction and given reactions

Target reaction: \(WO_3(s)+3H_2(g)\to W(s)+3H_2O(g)\)
Given reactions:

1. \(2W(s)+3O_2(g)\to 2WO_3(s)\), \(\Delta H^{\circ}_{rxn}=- 1685.8\space kJ/mol\)
2. \(2H_2(g)+O_2(g)\to 2H_2O(g)\), \(\Delta H^{\circ}_{rxn}=-483.6\space kJ/mol\)

Step2: Manipulate the given reactions

• Reverse reaction 1: \(2WO_3(s)\to 2W(s)+3O_2(g)\), \(\Delta H^{\circ}_{rxn}=+ 1685.8\space kJ/mol\)
• Multiply reaction 2 by \(\frac{3}{2}\): \(3H_2(g)+\frac{3}{2}O_2(g)\to 3H_2O(g)\), \(\Delta H^{\circ}_{rxn}=\frac{3}{2}\times(-483.6)= - 725.4\space kJ/mol\)

Step3: Add the manipulated reactions

\(2WO_3(s)\to 2W(s)+3O_2(g)\), \(\Delta H = + 1685.8\space kJ/mol\)
\(3H_2(g)+\frac{3}{2}O_2(g)\to 3H_2O(g)\), \(\Delta H=-725.4\space kJ/mol\)
Sum: \(2WO_3(s)+3H_2(g)\to 2W(s)+3H_2O(g)\)
\(\Delta H=(1685.8 - 725.4)\space kJ/mol = 960.4\space kJ/mol\)
Divide by 2 to get the target reaction:
\(WO_3(s)+3H_2(g)\to W(s)+3H_2O(g)\)
\(\Delta H=\frac{960.4}{2}=117.5\space kJ/mol\)

Answer:

\(117.5\space kJ/mol\)