Describe what is meant by the following:[2 marks]• Chemical energy• Heat (q)(ii) 500 mL of a gas is compressed to 250 mL under constant external pressure at3.00 atm. If the gas absorbs 12.6 kJ, what are the values of heat (q), work (w)and ΔE for the gas in kJ?[5 marks]What is the value of ΔE for the surroundings?Note: 1 atm = 101,325 Pa (N m-2)b. During volcanic eruptions, hydrogen sulfide gas is given off and oxidised byair according to the following chemical equation:2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)Calculate the standard enthalpy change for the reaction above, given:3S(s) + 2H2O(g) → 2H2S(g) + SO2(g) ΔH° = 162.9 kJ mol-1S(s) + O2(g) → SO2(g) ΔH° = -306.5 kJ mol-1[4 marks]c. (i) Write equilibrium expressions (Kc) for each of the following reactions.CO(g) + 2H2(g) ⇌ CH3OH(g)CO(g) + H2O(g) ⇌ CO2(g) + H2(g)PCl3(g) + Cl2(g) ⇌ PCl5(g)[6 marks](ii) For the endothermic gaseous equilibrium,PCl5(g) ⇌ PCl3(g) + Cl2(g)what is the effect of the following on the reaction direction?• Increased temperature• Increased pressureExplain your reasoning in each case.[3 marks]

(i) • Chemical energy: This is the energy stored in the bonds of chemical compounds, such as molecules and atoms. It is released or absorbed during chemical reactions. • Heat (q): This is the form of energy transfer between systems or objects with different temperatures, flowing from the hotter system to the cooler one until thermal equilibrium is reached.

(ii) To find the values of heat (q), work (w), and ΔE for the gas:

1. Heat (q): Given that the gas absorbs 12.6 kJ, q = +12.6 kJ.

2. Work (w): Work done on the gas during compression can be calculated using the formula: $w = -P_{\text{ext}} \Delta V$ where $P_{\text{ext}}$ is the external pressure and $\Delta V$ is the change in volume. $\Delta V = V_f - V_i = 250 \, \text{mL} - 500 \, \text{mL} = -250 \, \text{mL} = -0.250 \, \text{L}$ $w = - (3.00 \, \text{atm}) \times (-0.250 \, \text{L}) = 0.750 \, \text{L} \cdot \text{atm}$ Convert $\text{L} \cdot \text{atm}$ to kJ: $1 \, \text{L} \cdot \text{atm} = 101.325 \, \text{J}$ $w = 0.750 \, \text{L} \cdot \text{atm} \times 101.325 \, \text{J} / \text{L} \cdot \text{atm} = 75.99375 \, \text{J} = 0.07599375 \, \text{kJ} \approx 0.076 \, \text{kJ}$

3. Change in internal energy (ΔE): Using the first law of thermodynamics: $\Delta E = q + w$ $\Delta E = 12.6 \, \text{kJ} + 0.076 \, \text{kJ} = 12.676 \, \text{kJ}$

The value of ΔE for the surroundings is the negative of the ΔE for the system: $\Delta E_{\text{surroundings}} = -12.676 \, \text{kJ}$

(b) To calculate the standard enthalpy change (ΔH°) for the reaction: $2H_2S(g) + 3O_2(g) \rightarrow 2SO_2(g) + 2H_2O(g)$

Given: $3S(s) + 2H_2O(g) \rightarrow 2H_2S(g) + SO_2(g) \quad \Delta H° = 162.9 \, \text{kJ/mol}$ $S(s) + O_2(g) \rightarrow SO_2(g) \quad \Delta H° = -306.5 \, \text{kJ/mol}$

First, reverse the first reaction to match the desired products: $2H_2S(g) + SO_2(g) \rightarrow 3S(s) + 2H_2O(g) \quad \Delta H° = -162.9 \, \text{kJ/mol}$

Then, multiply the second reaction by 2: $2S(s) + 2O_2(g) \rightarrow 2SO_2(g) \quad \Delta H° = 2 \times (-306.5 \, \text{kJ/mol}) = -613.0 \, \text{kJ/mol}$

Add the two reactions: $2H_2S(g) + SO_2(g) + 2S(s) + 2O_2(g) \rightarrow 3S(s) + 2H_2O(g) + 2SO_2(g)$

Simplify: $2H_2S(g) + 3O_2(g) \rightarrow 2SO_2(g) + 2H_2O(g)$

Add the enthalpy changes: $\Delta H° = -162.9 \, \text{kJ/mol} + (-613.0 \, \text{kJ/mol}) = -775.9 \, \text{kJ/mol}$

(c) (i) Write equilibrium expressions (Kc):

1. $\text{CO(g) + 2H}_2\text{(g) ⇌ CH}_3\text{OH(g)}$ $K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}$

2. $\text{CO(g) + H}_2\text{O(g) ⇌ CO}_2\text{(g) + H}_2\text{(g)}$ $K_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]}$

3. $\text{PCl}_3\text{(g) + Cl}_2\text{(g) ⇌ PCl}_5\text{(g)}$ $K_c = \frac{[\text{PCl}_5]}{[\text{PCl}_3][\text{Cl}_2]}$

(ii) For the endothermic gaseous equilibrium: $\text{PCl}_5\text{(g) ⇌ PCl}_3\text{(g) + Cl}_2\text{(g)}$

• Increased temperature: The reaction will shift to the right (towards the products) because increasing temperature favors the endothermic direction, absorbing the added heat.

• Increased pressure: The reaction will shift to the left (towards the reactants) because increasing pressure favors the side with fewer moles of gas. In this case, the reactant side has fewer moles (1 mole of PCl₅ vs. 2 moles of products).