The following reaction is assumed to be in equilibrium:
\(2N{O_2}(g) \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {N_2}{O_4}(g)\)
If it's feasible to tell if the reaction started with pure sugar, that's great \(N{O_2}\)or with nothing but pure \({N_2}{O_4}\) it is necessary to explain whether or not this is the case.
The forward and backward (reverse) processes have identical rates in the equilibrium mixture.
For the following chemical reaction:
\(A \to B + C\)
The equilibrium constant will be expressed as:
\({K_C} = \frac{{(B\mid (C)}}{{(A)}}\)
The equilibrium concentrations of A, B, and C are (A), (B), and (C), respectively.
For the response:
\(2N{O_2}(g) \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {N_2}{O_4}(g)\)
Equilibrium constant for the aforementioned reaction is defined mathematically as:
\(Keq. = \frac{{\left. {\mid {N_2}{O_4}} \right)}}{{\left( {{{\left. {N{O_2}} \right|}^2}} \right.}}\)
As a result, we may state that when both reactants and products are accessible, the equilibrium constant has a specified value.
It no longer matters from which side the reaction begins because the Equilibrium value is constant (for a specific reaction at a constant temperature) and the reaction will automatically reach Equilibrium after a specified amount of time. The concentrations of reactants and products will become constant at the point of equilibrium.
And it's impossible to tell whether a reaction originates from pure energy or not if looking at it from the point where it reaches equilibrium \(N{O_2}\)or with pure \({N_2}{O_4}\)
As a result, it's impossible to say if a reaction starts from scratch \(N{O_2}\) or with pure \({N_2}{O_4}\).
