The graph below shows concentration changes of reactants and products for this reaction over time. Use this information to answer the following questions:

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

The reaction is at equilibrium during the intervals 0 to $t_1$, $t_1$ to $t_2$, and $t_3$ to $t_4$.

During the interval $t_2$ to $t_3$, the concentration of reactants ($[CO]$ and $[O_2]$) decreases, and the concentration of the product ($[CO_2]$) increases. This indicates that the forward reaction rate is faster.

At $t_4$, the temperature is decreased. The graph shows that after this change, the concentrations of $CO$ and $O_2$ decrease, while the concentration of $CO_2$ increases. This means the equilibrium shifts to the right (forward reaction). According to Le Chatelier's principle, a decrease in temperature favors the exothermic reaction. Therefore, the forward reaction is exothermic.

When the temperature is decreased at $t_4$, the system shifts to counteract this change by favoring the reaction that releases heat. Since the equilibrium shifts to the right, increasing the concentration of $CO_2$ and decreasing the concentrations of $CO$ and $O_2$, the forward reaction must be exothermic, as it releases heat.

At $t_2$, there is an instantaneous increase in the concentrations of all reactants and products. This indicates a decrease in the volume of the container (or an increase in pressure).

The change made at $t_2$ (a change in pressure/volume) does not affect the value of $K_c$. $K_c$ is only influenced by temperature. Therefore, $K_c$ will remain the same.