Question

sro(s)+co₂(g)→srco₃(s) δh°=-234kj/mol
2sr(s)+o₂(g)→2sro(s) δh°=-1184kj/mol
co₂(g)→c(s)+o₂(g) δh°=+394kj/mol
determine the molar enthalpy (in kj/mol) for the reaction
2srco₃(s)→2sr(s)+2c(s)+3o₂(g)

Explanation:

Step1: Manipulate the given equations

• Multiply the first equation \(SrO(s)+CO_{2}(g)\to SrCO_{3}(s)\), \(\Delta H^{\circ}=- 234\space kJ/mol\) by \(2\) and reverse it: \(2SrCO_{3}(s)\to2SrO(s) + 2CO_{2}(g)\), \(\Delta H_{1}=+468\space kJ/mol\)
• Reverse the second equation \(2Sr(s)+O_{2}(g)\to2SrO(s)\), \(\Delta H^{\circ}=-1184\space kJ/mol\): \(2SrO(s)\to2Sr(s)+O_{2}(g)\), \(\Delta H_{2}= + 1184\space kJ/mol\)
• Multiply the third equation \(CO_{2}(g)\to C(s)+O_{2}(g)\), \(\Delta H^{\circ}=+394\space kJ/mol\) by \(2\): \(2CO_{2}(g)\to2C(s)+2O_{2}(g)\), \(\Delta H_{3}=+788\space kJ/mol\)

Step2: Apply Hess's Law

Add the three manipulated equations:

$$
LATEXBLOCK0
$$

$$ \Delta H=(468 + 1184+788)\space kJ/mol $$

$$ \Delta H = 2440\space kJ/mol $$

Answer:

\(2440\space kJ/mol\)