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: $\Delta G_{rxn}=130.19J$

Explanation:

The balanced chemical reaction is,

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

The expression for Gibbs free energy change is,

$\Delta G_{rxn}=\sum [n\times \Delta G_(product)]-\sum [n\times \Delta G_(reactant)]$

$\Delta G_{rxn}=[(n_{CO_2}\times \Delta G_{CO_2})+(n_{CaO}\times \Delta G_{CaO})]-[(n_{CaCO_3}\times \Delta G_{CaCO_3})]$

where,

n = number of moles

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

$\Delta G_{rxn}=[(1\times -394.4)+(1\times -604.17)]-[(1\times -1128.76)]$

$\Delta G_{rxn}=130.19J$

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

Answer 2

Answer: 130 kJ

Explanation: