Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reaction CH3OH(g) + C0(g) → HCH,CO2(l) Round your answer to 2 significant digits. K= x10

[tex]\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}][/tex]

where,

[tex]\Delta G^o[/tex] = Gibbs free energy of reaction = ?

n = number of moles

[tex]\Delta G^0_{(HCH_3CO_2(g))}[/tex] = -389.8 kJ/mol

[tex]\Delta G^0_{(CH_3OH(g))}[/tex] = -161.96 kJ/mol

[tex]\Delta G^0_{(CO(g))}[/tex] = -137.2 kJ/mol

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

[tex]\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)][/tex]

[tex]\Delta G^o=-89.4kJ/mol[/tex]

The relation between the equilibrium constant and standard Gibbs, free energy is:

[tex]\Delta G^o=-RT\times \ln K[/tex]

where,

[tex]\Delta G^o[/tex] = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = [tex]30.0^oC=273+30.0=303K[/tex]

[tex]K[/tex] = equilibrium constant = ?

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

[tex]-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K[/tex]

[tex]K=2.6\times 10^{15}[/tex]

Thus, the value of [tex]K[/tex] for this reaction is, [tex]2.6\times 10^{15}[/tex]