Question

Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) Kp=2.26×104 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under each of the following conditions. PCH3OH= 1.4 atm ; PCO=PH2= 1.2×10−2 atm

Answer 1

Answer : The value of $\Delta G_{rxn}$ is, $8.867kJ/mole$

Explanation :

The formula used for $\Delta G_{rxn}$ is:

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$   ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction

$\Delta G_^o$ =  standard Gibbs free energy

R = gas constant = 8.314 J/mole.K

T = temperature = $25^oC=273+25=298K$

Q = reaction quotient

First we have to calculate the $\Delta G_^o$.

Formula used :

$\Delta G^o=-RT\times \ln K_p$

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

$\Delta G^o=-(8.314J/mole.K)\times (298K)\times \ln (2.26\times 10^{4})$

$\Delta G^o=-24839.406J/mole=-24.83\times 10^3J/mole=-24.83kJ/mole$

Now we have to calculate the value of 'Q'.

The given balanced chemical reaction is,

$CO(g)+2H_2(g)\rightarrow CH_3OH(g)$

The expression for reaction quotient will be :

$Q=\frac{(p_{CH_3OH})}{(p_{CO})\times (p_{H_2})^2}$

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

$Q=\frac{(1.4)}{(1.2\times 10^{-2})\times (1.2\times 10^{-2})^2}$

$Q=8.1\times 10^{5}$

Now we have to calculate the value of $\Delta G_{rxn}$ by using relation (1).

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$

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

$\Delta G_{rxn}=-24.83kJ/mole+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (8.1\times 10^{5})$

$\Delta G_{rxn}=8.867\times 10^3J/mole=8.867kJ/mole$

Therefore, the value of $\Delta G_{rxn}$ is, $8.867kJ/mole$