Question

Please answer this correctly cause there is no point of answering it if you are not sure you just confusing me more

Answer 1

First, calculate $\Delta G_{rxn}$ :

$BrCl\rightarrow \frac{1}{2}Br_{2(g)}+\frac{1}{2}Cl_{2(g)}$

$\Delta G_{rxn}=\frac{1}{2}\Delta G_{Br_2}+\frac{1}{2}\Delta G_{Cl_2}-\Delta G_{BrCl}$

$\Delta G_{rxn}=\frac{1}{2}*3.11+\frac{1}{2}*0-(-0.98)=2.535\frac{kJ}{mol}$

Now, the relationship between gibbs free energy and the equilibrium constant is:

$\Delta G_{rxn}=-RTLn(K_{eq})$

So:

$2535\frac{J}{mol}=-8.314\frac{J}{mol\cdot K}*298K*Ln(K_{eq})$

$-1.023=Ln(K_{eq})$

$K_{eq}=0.3595$

Now, we can calculate the concentrations at equilibrium:

$\\ BrCl\rightarrow \frac{1}{2}Br_{2(g)}+\frac{1}{2}Cl_{2(g)}\\ 1.3.............0..............0\\ -2x...........x.............x\\ 1.3-2x.......x..............x$

$K_{eq}=\frac{\frac{x}{12}*\frac{x}{12}}{\frac{1.3-2x}{12}}=0.3595$

$\frac{x^2}{1.3-2x}=4.314$

$x^2+8.628x-5.6082=0$

$x=0.607$

So:

$n_{BrCl}=1.3-2*0.607=0.086$

$n_{Br_2}=0.607$

$n_{Cl_2}=0.607$