Question

Calculate equilibrium constant (in multiples of \[{{10}^{-20}}\] ) when sodium reduces Aluminium oxide to aluminium at \[298\text{ }K.\] $(\Delta G_{f}^{0}$ of $N{{a}_{2}}{{O}_{3(s)}}$ at $298K=-377KJ/mole$ and $\Delta G_{f}^{0}$ of $A{{l}_{2}}{{O}_{3}}$ at $298K=-1582KJ/mole).$

Answer 1

Hint: We know that the equilibrium constant is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients. The ratio of concentration of products and reactants at equilibrium will be the required equilibrium constant.

Complete answer:
As we know that the sodium metal aluminate here, Meta refers to least hydrated form because sometimes it exists in the hydrated form in an excess of water containing environment. Exothermic reactions are the reactions which release energy in the form of heat and light and the reason for this energy release is because the total energy of the products is less than the total energy produced by the reactants. It is used to solidify the concrete, mainly when working during frost.
For a chemical reaction, the equilibrium constant can be defined as the ratio between the amount of reactant and the amount of product which is used to determine chemical behaviour. At equilibrium, Rate of the forward reaction $=$ Rate of the backward reaction. Thus, the chemical reaction is given by;
$A{{l}_{2}}{{O}_{3}}+2{{N}_{4}}{{O}_{4}}\to 2{{N}_{4}}Al{{O}_{2}}+{{H}_{2}}O.$
The equilibrium constant of a chemical reaction usually denoted by the symbol K provides insight into the relationship between the products and reactants when a chemical reaction reaches equilibrium. For example, the equilibrium constant of concentration of a chemical reaction at equilibrium can be defined as the ratio of the concentration of products to the concentration of the reactants, each raised to their respective stoichiometric coefficients.
Now, $\Delta G=product-Reactant=2\left( -377 \right)-\left( -1582 \right)=-754+1582=828.$
We have formula $\Delta G=-RT\ln (Eq)$ on substitution we get; $828=-8.314\times 298\times \log \left( Eq \right)$
On further solving we get; \[-0.145=\log \left( Eq \right)\Rightarrow 0.716=Eq\]

Note:
Remember that the aluminium and that is sodium hydroxide cannot be stored in the same containers because they both react heavily with the evolution of heat and form sodium aluminate as a product of reaction. Ortho, Meta, Para, etc., are all variations of something called the condensation degree of the aluminate ions.