Given the data in Appendix B in the textbook, calculate Kp at 25 ∘C for the reaction MgCO3(s)⇌MgO(s)+CO2(g) lnK = -19.45...

Question

Question

Given the data in Appendix B in the textbook, calculate Kp at 25 ∘C for the reaction
MgCO3(s)⇌MgO(s)+CO2(g)

lnK = -19.45

Answer 1

Answer #1

ΔG⁰_rxn = ∑[ΔG_f^o(products)] - ∑[ΔG_f^o(reactants)]

ΔG⁰_rxn = [-569.4 -394.4] - [-1012.1]

ΔG⁰_rxn = 48.3 kJ/mol = 48300 J/mol

ΔG⁰_rxn = -RTlnKp; where R = 8.3145 J/mol.K; T = 25 + 273.15 = 298.15 K

lnKp = -ΔG⁰_rxn / RT = -48300 J/mol / (8.3145 J/mol.K x 298.15 K) = -19.4839128986

lnKp = -19.4839128986

Kp = e^{-19.4839128986} = 3.45337819 x 10^-9

Kp at 25 ⁰C for the reaction = 3.453 x 10^-9

In scientific notation answer = 3.453E-09

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Answer 2

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