Explain chemical equilibrium and determine the heat of reaction for a given scenario using Le Chatelier's principle.

At $300 \text{ K}$, the moles of $P(g)$ are $0.8 \text{ mol}$. When the temperature is increased to $350 \text{ K}$, the moles of $P(g)$ increase to $1.2 \text{ mol}$. Since $P(g)$ is a reactant, an increase in its amount indicates that the reverse reaction is favored.

According to Le Chatelier's principle, an increase in temperature favors the endothermic reaction. Since the reverse reaction ($2PQ(g) \rightarrow 2P(g) + Q_2(g)$) is favored by an increase in temperature, the reverse reaction is endothermic. Therefore, the forward reaction ($2P(g) + Q_2(g) \rightleftharpoons 2PQ(g)$) is exothermic.
The heat of the reaction ($\Delta H$) for the forward reaction is NEGATIVE.

6.2.2
Le Chatelier's principle states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress. In this case, increasing the temperature is a stress. The system relieves this stress by favoring the reaction that absorbs heat, which is the endothermic reaction. Since the amount of reactant $P(g)$ increased, the equilibrium shifted to the left (reverse reaction). This means the reverse reaction is endothermic, and consequently, the forward reaction is exothermic.

6.2.3

The initial moles at $300 \text{ K}$ are:
$n_P = 0.8 \text{ mol}$
$n_{Q_2} = 0.8 \text{ mol}$
$n_{PQ} = 3.2 \text{ mol}$
At $350 \text{ K}$, the new equilibrium has $n_P = 1.2 \text{ mol}$.
The change in moles of $P$ is $\Delta n_P = 1.2 \text{ mol} - 0.8 \text{ mol} = +0.4 \text{ mol}$.
Since the reaction is $2P(g) + Q_2(g) \rightleftharpoons 2PQ(g)$, and $P$ increased by $0.4 \text{ mol}$, the reverse reaction occurred.
Based on the stoichiometry of the reverse reaction ($2PQ(g) \rightarrow 2P(g) + Q_2(g)$):

• If $P$ increased by $0.4 \text{ mol}$, then $Q_2$ increased by $\frac{1}{2} \times 0.4 \text{ mol} = 0.2 \text{ mol}$.
• $PQ$ decreased by $\frac{2}{2} \times 0.4 \text{ mol} = 0.4 \text{ mol}$. The moles at $350 \text{ K}$ equilibrium are: $n_P = 0.8 \text{ mol} + 0.4 \text{ mol} = 1.2 \text{ mol}$ $n_{Q_2} = 0.8 \text{ mol} + 0.2 \text{ mol} = 1.0 \text{ mol}$ $n_{PQ} = 3.2 \text{ mol} - 0.4 \text{ mol} = 2.8 \text{ mol}$

The volume of the container is $2 \text{ dm}^3$.
$[P] = \frac{1.2 \text{ mol}}{2 \text{ dm}^3} = 0.6 \text{ M}$
$[Q_2] = \