The thermodynamic values from part A will be useful as you work through part B:

ΔH∘rxn 178.5kJ/mol
ΔS∘rxn 161.0J/(mol⋅K)
Calculate the equilibrium constant for the following reaction at room temperature, 25 ∘C :

CaCO3(s)→CaO(s)+CO2(g)

1.33 × 10⁻²³

Explanation:

Let's consider the following equation.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression.

ΔG°rxn = ΔH°rxn - T × ΔS°rxn

where,

ΔG°rxn = 178.5 kJ/mol - 298 K × 0.1610 kJ/mol.K

ΔG°rxn = 130.5 kJ/mol

Then, we can calculate the equilibrium constant (K) using the following expression.

ΔG°rxn = - R × T × ln K

where,

ΔG°rxn = - R × T × ln K

ln K = -ΔG°rxn / R × T

ln K = -(130.5 × 10³ J/mol) / (8.314 J/mol.K) × 298 K

K = 1.33 × 10⁻²³

explanation: it gains negative electrons, therefore becoming negatively charged. as charge cannot be created, nor destroyed, only transferred, this leaves the wool positive. now that they have opposite charges, the two items attract.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.