Question

given these reactions, where x represents a generic metal or metalloid

1. h₂(g) + ½ o₂(g) → h₂o(g) δh₁ = -241.8 kj
2. x(s) + 2cl₂(g) → xcl₄(s) δh₂ = +227.7 kj
3. ½ h₂(g) + ½ cl₂(g) → hcl(g) δh₃ = -92.3 kj
4. x(s) + o₂(g) → xo₂(s) δh₄ = -555.3 kj
5. h₂o(g) → h₂o(l) δh₅ = -44.0 kj

what is the enthalpy, δh, for this reaction?
xcl₄(s) + 2h₂o(l) → xo₂(s) + 4hcl(g)

Explanation:

Step1: Reverse reaction 2

$$\ce{XCl4(s) \to X(s) + 2Cl2(g)}\quad \Delta H = - 227.7\,\text{kJ}$$

Step2: Reverse reaction 5 and multiply by 2

$$\ce{2H2O(l)\to 2H2O(g)}\quad \Delta H = 88.0\,\text{kJ}$$

Step3: Multiply reaction 3 by 4

$$\ce{2H2(g) + 2Cl2(g)\to 4HCl(g)}\quad \Delta H = - 369.2\,\text{kJ}$$

Step4: Use reaction 4

$$\ce{X(s) + O2(g)\to XO2(s)}\quad \Delta H = - 555.3\,\text{kJ}$$

Step5: Add all the reactions and their enthalpies

$$\begin{align*} \ce{XCl4(s) + 2H2O(l)}&\ce{\to XO2(s) + 4HCl(g)}\\ \Delta H&=(-227.7 + 88.0-369.2 - 555.3)\,\text{kJ}\\ &=(-227.7-369.2-555.3 + 88.0)\,\text{kJ}\\ &=(-1152.2+88.0)\,\text{kJ}\\ &=-1064.2\,\text{kJ} \end{align*}$$

Answer:

\(-1064.2\,\text{kJ}\)