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Yes the system is at equilibrium if reactants and products are at equilibrium at 10⁻³ < K <10³.\
When opposing reactions are happening at the same pace, chemical equilibrium happens.
[tex]N_2O_4 (g)[/tex] → [tex]2 NO _2(g)[/tex]
We represent the system's equation with a double arrow since the forward and reverse processes are both active in an equilibrium state.
Forward Reaction: Rate Law:
[tex]N_2O_4(g)[/tex] → [tex]2 NO_2 (g)[/tex] Rate = [tex]k_f [N_2O_4][/tex]
Reverse Reaction: Rate Law:
[tex]2 NO_2 (g)[/tex] → [tex]N_2O_4(g)[/tex] Rate = [tex]k_f [NO_2]^2[/tex]
Therefore at equilibrium;
[tex]Rate_f=Rate_r\\[/tex]
[tex]k_f[N_O_4] = k_r[NO_2]^2[/tex]
It becomes:
[tex]\frac{k_f}{k_r} = \frac{[NO_2]^2}{[N_2O_4]} \\[/tex]
[tex]K_e_q = \frac{k_f}{k_r} = \frac{[NO_2]^2}{[N_2O_4]} = a constant[/tex]
The reciprocal of the equilibrium constant of the forward reaction is the equilibrium constant of a reaction moving in the opposite direction.
The reaction is said to have a large amount of both reactants and products at equilibrium at 10⁻³ < K <10³.
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