Question

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant k at 25.0 celsius for the following reaction.
6Cl2(g)+2Fe2O3(s)----->4FeCl3(s)+3O2
Round answer to 2 significant digits.

Answer 1

Answer:

Explanation:

From the given reaction:

$6Cl_{2(g)}+2Fe_2O_{3(s)} \to 4FeCl_{3(s)}+3O_2$

From the Gibbs Free Energy table at standard conditions, the value of each compound is as follows:

$G_f^0 \ of \ Cl_2 = 0 \ KJ/mol$                       $G_f^0 \ of \ Fe_2O_3 = -742.24 \ KJ/mol$

$G_f^0 \ of \ Fe_2Cl_3 = -334.05 \ KJ/mol$      $G_f^0 \ of \ O_2 = 0 \ KJ/mol$

Now, the standard Gibb's Free energy for the given reaction can be estimated as follows:

$\mathtt{\Delta G^0 = (4 *G_f^0(FeCl_3) +3*G_f^0(O_2)) - (6*G_f^0 (Cl_2) +2*G_f^0(Fe_2O_3))}$

$\mathtt{\Delta G^0 = (4 *(-334.05) +3*(0)) - (6(0) +2(-742.24))}$

$\mathtt{\Delta G^0 = 148.28 \ kJ/mol}$

using the following formula:

$\mathtt{\Delta G^0 =-RTIn K_{eq}}$

the equilibrium constant can  be determined as:

$\mathtt{ In K_{eq} =\dfrac{\Delta G^0 }{-RT}}$

$\mathtt{ In K_{eq} =\dfrac{148.28*10^3 J/mol }{-(8.314 \ J/k mol )*298 \ K}}$

$\mathtt{ In K_{eq} =-59.85}$

$\mathtt{ K_{eq} =e^{-59.85}}$

$\mathtt{ K_{eq} =1.0*10^{-26}}$ to 2 significant figures.