Recall how to find the expression of the equilibrium constant for the simplified equation:

$\displaystyle aA+bB\rightleftharpoons cC+dD$

$\displaystyle \text{K}_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

Since the given equation has gases, we will only consider the partial pressures of each gas in the expression for the equilibrium constant. Remember that only molecules in aqueous and gas forms are included in this expression. Pure solids and pure liquids are excluded.

Thus, we can then write the following equilibrium constant for the given equation:

$\displaystyle \text{K}_{eq}=\frac{(\text{P}_{SiO_2})^4}{(\text{P}_{Si_2H_6})^2(\text{P}_{O_2})^7}$

Start by writing the equilibrium expression:

$\displaystyle K=\frac{[Cl_2]^2}{[HCl]^4[O_2]}$

Now, create a chart like the following to keep track of the changes in concentration.

	$\displaystyle [HCl]$	$\displaystyle [O_2]$	$\displaystyle [Cl_2]$
Initial	0.750	2.00	0.00
Change	-0.100	-0.025	0.200
Equilibrium	0.650	19.75	0.200

Since we know that the concentration of HCl decreased by $\displaystyle 0.100M$, we can use the stoichiometric ratios to deduce the amount of change for the oxygen gas and the chlorine gas.

Plug in the equilibrium concentrations into the expression for the equilibrium constant.

$\displaystyle K=\frac{(0.200)^2}{(0.650)^4(1.975)}=0.113$

Start by using the free energies of formation to find $\displaystyle \Delta G^{\degree}_{rxn}$.

 $\displaystyle \Delta G^{\degree}_{rxn}=\Sigma n\Delta G^{\degree}_f \text{ of products}-\Sigma n\Delta G^{\degree}_f \text{ of reactants}$, 

$\displaystyle \Delta G^{\degree}_{rxn}=2(-5.44)-19.3=-30.18\text{ kJ}$

Recall the equation that links together $\displaystyle \Delta G^{\degree}_{rxn}$ with the equilibrium constant, $\displaystyle K$.

$\displaystyle \Delta G^{\degree}_{rxn} =-RT \ln K$

Plug in the given information and solve for $\displaystyle K$.

$\displaystyle -30.18(1000)=-(8.314)(298)\ln K$

$\displaystyle \ln K=12.18$

$\displaystyle K=e^{12.18}=1.95\times 10^5$

$\displaystyle k_{p}= \frac{P(product)^{a}}{P(reactant)^{b}*P(reactant)^{c}}$

$\displaystyle k_{p}=30.6= \frac{P(NOBr)^{2}}{(\frac{132}{760})*(\frac{113}{760})^{2}}$

Use algebra to solve for the partial pressure of $\displaystyle NOBr$.

$\displaystyle \sqrt{30.6*(\frac{132}{760})*(\frac{113}{760})^{2}}=P(NOBr)$ 

$\displaystyle 0.343atm=P(NOBr)$

$\displaystyle Q_{p}=\frac{(\frac{125}{760})^{2}}{(\frac{125}{760})*(\frac{125}{760})^{2}}=6.08$

$\displaystyle k_p=30.6$

Since $\displaystyle k>Q$ the reaction is not at equilibrium. This means that at equilibrium, the ratio of products to reactants is greater than at the given conditions. Thus, the reaction will move right towards the products to reach equilibrium.

Use an ice table and the $\displaystyle k_{c}$ equation to solve.                      

                        $\displaystyle CO$             $\displaystyle H_{2}O$            $\displaystyle CO_{2}$              _{$\displaystyle H_{2}$} 

Initial              $\displaystyle .43M$          $\displaystyle .48M$              $\displaystyle 0$                 $\displaystyle 0$

Change        $\displaystyle -.19M$      $\displaystyle -.19M$        $\displaystyle +.19M$      $\displaystyle +.19M$   

Equilibrium      $\displaystyle .24M$          $\displaystyle .29M$           $\displaystyle .19M$         $\displaystyle .19M$    

$\displaystyle k_{c}= \frac{[H_{2}][CO_{2}]}{[CO][H_{2}O]}$

$\displaystyle k_{c}=\frac{.19M*.19M}{.24M*.29M}=.52$

Use an ICE table and the $\displaystyle k_{c}$ equation to solve.

                         $\displaystyle [N_{2}O_{4}]$     $\displaystyle [NO_{2}]$ 

Initial               $\displaystyle .027M$         $\displaystyle 0M$        

Change         $\displaystyle -.012M$    $\displaystyle +.012M$   

Equilibrium      $\displaystyle .015M$       $\displaystyle .012M$  

$\displaystyle k_{c}= \frac{[NO_{2}]^2}{[N_{2}O_{4}]}=\frac{.012M^2}{.015M}=.0096$

For this question, we're given the acid dissociation constant for a reaction that occurs at a given temperature. We're asked to find the standard free energy change for the reaction.

First, we're going to need to use an equation that relates standard free energy changes with an equilibrium constant, which is shown as follows.

$\displaystyle \Delta G^{o}=-RTln(K)$

With regards to the temperature, we will need to convert the units given in the question stem into units of Kelvin.

$\displaystyle T=25+273=298\:K$

Knowing that $\displaystyle R$ is the ideal gas constant, we have all the information we need to solve for the value of $\displaystyle \Delta G^{o}$.

$\displaystyle \Delta G^{o}=-(8.314\:\frac{J}{mol\cdot K})(298\:K)ln(2.3\cdot 10^{-9})$

$\displaystyle \Delta G^{o}=49279\:\frac{J}{mol}\approx 49\:\frac{kJ}{mol}$

Acid dissociation constant which is denoted as $\displaystyle K_{a}$ is the equilibrium constant for the ionization of an acid. Therefore, the numerator contains the product of the concentrations of the substances on the product side of the chemical equation. The denominator contains the product of the concentrations of the substances on the reactant side of the chemical equation. $\displaystyle H_{2}O$ is omitted in the acid dissociation constant expression because as the solvent it is in excess and therefore the change in its concentration is negligible in comparison to the other substances in solution.