Dinitrogen tetraoxide, N2O4(g), decomposes to nitrogen dioxide, NO2(g), in a sealed syringe of volume 2 dm3.

Dipping the syringe into ice water decreases the temperature. The brown colour disappears, meaning the concentration of $\text{NO}_2\text{(g)}$ (brown) decreases, and the equilibrium shifts to the left (towards the colourless $\text{N}_2\text{O}_4\text{(g)}$).

According to Le Chatelier's principle, a decrease in temperature favors the exothermic reaction. Since the equilibrium shifted to the left, the reverse reaction is favored and must be exothermic.

If the reverse reaction is exothermic, then the forward reaction must be ENDOTHERMIC.

6.3
Decreasing the volume of the syringe increases the pressure. The equilibrium reaction is $\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)}$. There is 1 mole of gas on the reactant side and 2 moles of gas on the product side. An increase in pressure shifts the equilibrium to the side with fewer moles of gas, which is the reactant side (left).

6.3.1 The number of moles of $\text{N}_2\text{O}_4\text{(g)}$
Since the equilibrium shifts to the left, more $\text{N}_2\text{O}_4\text{(g)}$ is formed.
The number of moles of $\text{N}_2\text{O}_4\text{(g)}$ INCREASES.

6.3.2 The value of the equilibrium constant
The equilibrium constant ($K_c$) only changes with temperature. Since the temperature is kept constant, $K_c$ REMAINS THE SAME.

6.3.3 The rate of the forward and reverse reactions
Decreasing the volume increases the concentration of both reactants and products. Higher concentrations lead to more frequent collisions, thus increasing the rate of both forward and reverse reactions.
The rate of the forward and reverse reactions INCREASES.

6.4

The reaction is:

Initial moles:
$\text{N}_2\text{O}_4 = X$
$\text{NO}_2 = 0$

$20\%$ of $\text{N}_2\text{O}_4$ decomposed.
Moles of $\text{N}_2\text{O}_4$ decomposed $= 0.20X$
Moles of $\text{NO}_2$ formed $= 2 \times (\text{moles of N}_2\text{O}_4 \text{ decomposed}) = 2 \times 0.20X = 0.40X$

Equilibrium moles of $\text{N}_2\text{O}_4 = X - 0.20X = 0.80X$
Equilibrium moles of $\text{NO}_2 = 0 + 0.40X = 0.40X$

Volume of syringe, $V = 2 \text{ dm}^3$.

Given $K_c = 0.16$.