Question

Calculate AGrxn for this equation, rounding your

answer to the nearest whole number.

CaCO3(s)– Cao(s) + CO2(g)

AGf.Cacoa = -1,128.76 kJ/mol

AGf, Cao = -604.17 kJ/mol

AGT,CO, = -394.4 kJ/mol

AGrx = what

Answer 1

Answer: 130 kJ

Step-by-step explanation:

Answer 2

Answer:
`\Delta G_(rxn)=130.19J`

Step-by-step explanation:

The balanced chemical reaction is,

`CaCO_3(s)\rightarrow CaO(s)+CO_2(g)`

The expression for Gibbs free energy change is,

`\Delta G_(rxn)=\sum [n* \Delta G_(product)]-\sum [n* \Delta G_(reactant)]`

`\Delta G_(rxn)=[(n_(CO_2)* \Delta G_(CO_2))+(n_(CaO)* \Delta G_(CaO))]-[(n_(CaCO_3)* \Delta G_(CaCO_3))]`

where,

n = number of moles

Now put all the given values in this expression, we get

`\Delta G_(rxn)=[(1* -394.4)+(1* -604.17)]-[(1* -1128.76)]`

`\Delta G_(rxn)=130.19J`

Therefore, the gibbs free energy for this reaction is, +130.19 kJ