1 Chemical equilibrium is a dynamic state in a reversible chemical reaction where the rate of the forward reaction is equal to the rate of the reverse reaction, resulting in constant concentrati

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6.1 Chemical equilibrium is a dynamic state in a reversible chemical reaction where the rate of the forward reaction is equal to the rate of the reverse reaction, resulting in constant concentrations of reactants and products.

6.2.1 At $t_1$, the concentrations of $N_2O_4(g)$ are decreasing and $NO_2(g)$ are increasing, indicating that the system is not yet at equilibrium and the forward reaction is dominant.
The rate of the forward reaction is HIGHER THAN the rate of the reverse reaction.

6.2.2 At $t_2$, the concentrations of both $N_2O_4(g)$ and $NO_2(g)$ have become constant, which signifies that the system has reached equilibrium.
The rate of the forward reaction is EQUAL TO the rate of the reverse reaction.

6.3.1 At $t_3$, there is an instantaneous decrease in the concentrations of both $N_2O_4(g)$ and $NO_2(g)$, followed by a shift in equilibrium where $[N_2O_4]$ decreases further and $[NO_2]$ increases. Since the equilibrium constant ($K_c$) did not change, the temperature remained constant. An instantaneous decrease in concentration for all species indicates an increase in volume (or decrease in pressure). The equilibrium shifts to the side with more moles of gas ($2NO_2(g)$) to counteract the pressure decrease.
The change made was an increase in volume (or decrease in pressure).

6.3.2 At $t_4$, there is an instantaneous increase in the concentrations of both $N_2O_4(g)$ and $NO_2(g)$, followed by a shift in equilibrium where $[N_2O_4]$ increases further and $[NO_2]$ decreases. Since $K_c$ did not change, the temperature remained constant. An instantaneous increase in concentration for all species indicates a decrease in volume (or increase in pressure). The equilibrium shifts to the side with fewer moles of gas ($N_2O_4(g)$) to counteract the pressure increase.
The change made was a decrease in volume (or increase in pressure).

6.4 The reaction is $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ with $\Delta H > 0$.