Question

using the equations
$\ce{sr(s) + c(s) + \frac{3}{2}o_{2}(g) \to srco_{3}(s)}$ $\delta h^{\circ}=-234\\ \text{kj/mol}$
$\ce{2sr(s) + o_{2}(g) \to 2sro(s)}$ $\delta h^{\circ}=-1184\\ \text{kj/mol}$
$\ce{co_{2}(g) \to c(s) + o_{2}(g)}$ $\delta h^{\circ}=+394\\ \text{kj/mol}$
determine the molar enthalpy (in kj/mol)
for the reaction
$\ce{2srco_{3}(s) \to 2sr(s) + 2c(s) + 3o_{2}(g)}$

Explanation:

Step1: Reverse equations and adjust $\Delta H^{\circ}$

• Reverse the first equation: \(2SrCO_{3}(s)\to 2Sr(s)+2C(s) + 3O_{2}(g)\)
• If \(Sr(s)+C(s)+\frac{3}{2}O_{2}(g)\to SrCO_{3}(s)\), \(\Delta H^{\circ}=- 234\space kJ/mol\), then \(2SrCO_{3}(s)\to 2Sr(s)+2C(s)+3O_{2}(g)\), \(\Delta H_{1}^{\circ}=2\times234\space kJ/mol\)
• Reverse the second equation: \(2SrO(s)\to 2Sr(s)+O_{2}(g)\), \(\Delta H_{2}^{\circ}=1184\space kJ/mol\)
• Reverse the third equation: \(CO_{2}(g)\to C(s)+O_{2}(g)\), \(\Delta H_{3}^{\circ}=- 394\space kJ/mol\)

Step2: Use Hess's Law

The target reaction \(2SrCO_{3}(s)\to 2Sr(s)+2C(s)+3O_{2}(g)\) can be obtained by:
\(2\times\) (reverse of \(Sr(s)+C(s)+\frac{3}{2}O_{2}(g)\to SrCO_{3}(s)\))+\(2\times\) (reverse of \(CO_{2}(g)\to C(s)+O_{2}(g)\))+(reverse of \(2SrO(s)\to 2Sr(s)+O_{2}(g)\))

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Answer:

\(864\space kJ/mol\)