Pure carbon dioxide (PCO2 = 1 atm) decomposes at high temperature. For the reaction system 2 〖CO〗_2 (g) ⇌2 CO(g)+ O_2 (g) Is this reaction endothermic? Calculate the value of Kp at each temperature.

The reaction decomposes more as T increases, therefore it is ENDOTHERMIC, meaning it requires energy to form CO and O₂.

Kp for each specie...

Kp = CO^2 O2 / (CO2)^2

for

T = 1500

Assume 1 atm for CO2, after % dissociation

P-CO2 left = 1*(1-0.048/100)= 0.99952

P-CO formed = 1-0.99952 = 0.00048

P-O2 = (1-0.99952)/2 = 0.00024

so..

Kp = CO^2 O2 / (CO2)^2

Kp = (0.00048^2)(0.00024) / (0.99952^2) = 5.53*10^-11

T = 2500

Assume 1 atm for CO2, after % dissociation

P-CO2 left = 1*(1-17.6/100)= 0.824

P-CO formed = 1-0.824= 0.176

P-O2 = (1-0.176)/2 = 0.088

so..

Kp = CO^2 O2 / (CO2)^2

Kp = (0.176^2)(0.088) / (0.824^2) =0.0040

T = 3000

Assume 1 atm for CO2, after % dissociation

P-CO2 left = 1*(1-54.8/100)= 0.452

P-CO formed = 1-0.452= 0.548

P-O2 = (1-0.452)/2 = 0.274

so..

Kp = CO^2 O2 / (CO2)^2

Kp = (0.548^2)(0.274) / (0.452^2) =0.4027

Explanation:

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

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explanation:

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