Respuesta :
Answer : The value of [tex]\Delta G_{rxn}[/tex] is, [tex]8.867kJ/mole[/tex]
Explanation :
The formula used for [tex]\Delta G_{rxn}[/tex] is:
[tex]\Delta G_{rxn}=\Delta G^o+RT\ln Q[/tex] ............(1)
where,
[tex]\Delta G_{rxn}[/tex] = Gibbs free energy for the reaction
[tex]\Delta G_^o[/tex] = standard Gibbs free energy
R = gas constant = 8.314 J/mole.K
T = temperature = [tex]25^oC=273+25=298K[/tex]
Q = reaction quotient
First we have to calculate the [tex]\Delta G_^o[/tex].
Formula used :
[tex]\Delta G^o=-RT\times \ln K_p[/tex]
Now put all the given values in this formula, we get:
[tex]\Delta G^o=-(8.314J/mole.K)\times (298K)\times \ln (2.26\times 10^{4})[/tex]
[tex]\Delta G^o=-24839.406J/mole=-24.83\times 10^3J/mole=-24.83kJ/mole[/tex]
Now we have to calculate the value of 'Q'.
The given balanced chemical reaction is,
[tex]CO(g)+2H_2(g)\rightarrow CH_3OH(g)[/tex]
The expression for reaction quotient will be :
[tex]Q=\frac{(p_{CH_3OH})}{(p_{CO})\times (p_{H_2})^2}[/tex]
In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.
Now put all the given values in this expression, we get
[tex]Q=\frac{(1.4)}{(1.2\times 10^{-2})\times (1.2\times 10^{-2})^2}[/tex]
[tex]Q=8.1\times 10^{5}[/tex]
Now we have to calculate the value of [tex]\Delta G_{rxn}[/tex] by using relation (1).
[tex]\Delta G_{rxn}=\Delta G^o+RT\ln Q[/tex]
Now put all the given values in this formula, we get:
[tex]\Delta G_{rxn}=-24.83kJ/mole+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (8.1\times 10^{5})[/tex]
[tex]\Delta G_{rxn}=8.867\times 10^3J/mole=8.867kJ/mole[/tex]
Therefore, the value of [tex]\Delta G_{rxn}[/tex] is, [tex]8.867kJ/mole[/tex]
