Respuesta :
Answer : The value of [tex]K_p[/tex] for this reaction is, [tex]2.73\times 10^{42}[/tex]
Explanation :
The given chemical reaction is:
[tex]C_2H_2(g)+2H_2(g)\rightarrow C_2H_6(g)[/tex]
Now we have to calculate value of [tex](\Delta G^o)[/tex].
[tex]\Delta G^o=G_f_{product}-G_f_{reactant}[/tex]
[tex]\Delta G^o=[n_{C_2H_6(g)}\times \Delta G^0_{(C_2H_6(g))}]-[n_{C_2H_2(g)}\times \Delta G^0_{(C_2H_2(g))}+n_{H_2(g)}\times \Delta G^0_{(H_2(g))}][/tex]
where,
[tex]\Delta G^o[/tex] = Gibbs free energy of reaction = ?
n = number of moles
[tex]\Delta G^0_{(C_2H_6(g))}[/tex] = -32.89 kJ/mol
[tex]\Delta G^0_{(C_2H_2(l))}[/tex] = 209.2 kJ/mol
Now put all the given values in this expression, we get:
[tex]\Delta G^o=[1mole\times (-32.89kJ/mol)]-[1mole\times (209.2kJ/mol)+2mole\times (0kJ/mol)][/tex]
[tex]\Delta G^o=-242.09kJ/mol[/tex]
The relation between the equilibrium constant and standard Gibbs, free energy is:
[tex]\Delta G^o=-RT\times \ln K_p[/tex]
where,
[tex]\Delta G^o[/tex] = standard Gibbs, free energy
R = gas constant = 8.314 J/L.atm
T = temperature = 298 K
[tex]K_p[/tex] = equilibrium constant = ?
Now put all the given values in this expression, we get:
[tex]-242.09kJ/mol=-(8.314J/L.atm)\times (298K)\times \ln K_p[/tex]
[tex]K_p=2.73\times 10^{42}[/tex]
Thus, the value of [tex]K_p[/tex] for this reaction is, [tex]2.73\times 10^{42}[/tex]
