Question

using the equations

sro(s)+co₂(g)→srco₃(s) δh°=

2sr(s)+o₂(g)→2sro(s) δh°=-1

co₂(g)→c(s)+o₂(g) δh°=+394

determine the molar enthalpy (in kj/mol) for the reaction

2srco₃(s)→2sr(s)+2c(s)+3o₂(g)

answer:

kj/mol

Explanation:

Step1: Reverse and scale equations

• Reverse the first equation: \(2SrCO_3(s)\to2SrO(s) + 2CO_2(g)\), \(\Delta H_1= 2\times(+114\ kJ)\)
• Reverse the second equation: \(2SrO(s)\to2Sr(s)+O_2(g)\), \(\Delta H_2 =+1184\ kJ\)
• Scale the third equation by 2: \(2CO_2(g)\to2C(s)+2O_2(g)\), \(\Delta H_3=2\times(+394\ kJ)\)

Step2: Add the enthalpy changes

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Answer:

\(2200\)