Question

Calculate ΔGrxn at 298 K under the conditions shown below for the following reaction.

Answer 1

Answer : The value of $\Delta G_{rxn}$ is -24.9 kJ/mol

Explanation :

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

The given balanced chemical reaction is,

$Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)$

The expression for reaction quotient will be :

$Q=\frac{(p_{CO_2})^3}{(p_{CO})^3}$

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

$Q=\frac{(2.1)^3}{(1.4)^2}$

$Q=3.375$

Now we have to calculate the value of $\Delta G_{rxn}$.

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  = -28.0 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

Q = reaction quotient = 3.375

Now put all the given values in the above formula 1, we get:

$\Delta G_{rxn}=(-28.0kJ/mol)+[(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln (3.375)$

$\Delta G_{rxn}=-24.9kJ/mol$

Therefore, the value of $\Delta G_{rxn}$ is -24.9 kJ/mol