Q.Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below :
2 Zn(s) + O2(g) → 2 ZnO(s) ; ΔH = -693.8 kJ mol^-1 (Note: more than one of the given options may be correct.)

The key idea here is the Enthalpy of Reaction and its interpretation.

The given reaction is $2 \text{ Zn(s)} + \text{O}_2\text{(g)} \rightarrow 2 \text{ ZnO(s)}$ with $\Delta H = -693.8 \text{ kJ mol}^{-1}$.

1. A negative value for $\Delta H$ indicates an exothermic reaction, meaning energy is released or evolved during the reaction. The magnitude $693.8 \text{ kJ}$ is evolved for the reaction as written (i.e., when 2 moles of Zn react with 1 mole of O$_2$).
2. For an exothermic reaction, the total enthalpy of the products is less than the total enthalpy of the reactants. This is because the system releases energy, resulting in products with lower energy content. $\Delta H = H_{\text{products}} - H_{\text{reactants}}$ Since $\Delta H < 0$, it implies $H_{\text{products}} < H_{\text{reactants}}$.
3. Comparing this understanding with the given options: …

In chemistry, the enthalpy change ($\Delta H$) for a reaction tells us about the heat absorbed or released during a chemical process at constant pressure. It's a crucial concept for understanding the energy balance of reactions.

The given reaction is:

$2 \text{ Zn(s)} + \text{O}_2\text{(g)} \rightarrow 2 \text{ ZnO(s)}$

with $\Delta H = -693.8 \text{ kJ mol}^{-1}$.

Let's break down what this $\Delta H$ value signifies.

Understanding Enthalpy Change ($\Delta H$)

1. Definition: Enthalpy change ($\Delta H$) is defined as the difference between the total enthalpy of the products and the total enthalpy of the reactants.

Here, $H_{\text{products}}$ refers to the total enthalpy of 2 moles of ZnO, and $H_{\text{reactants}}$ refers to the total enthalpy of 2 moles of Zn and 1 mole of O$_2$.

2. Sign Convention:

* A negative value of $\Delta H$ (as in this case, $-693.8 \text{ kJ mol}^{-1}$) indicates an exothermic reaction. This means that the products have lower enthalpy than the reactants, and the excess energy is released or evolved into the surroundings, usually as heat.

* A positive value of $\Delta H$ indicates an endothermic reaction. This means the products have higher enthalpy than the reactants, and energy is absorbed from the surroundings.

3. Units: The unit $\text{kJ mol}^{-1}$ means "kilojoules per mole of reaction as written." For the given reaction, it means $693.8 \text{ kJ}$ of energy is involved when 2 moles of Zn react with 1 mole of O$_2$ to form 2 moles of ZnO.

Now, let's evaluate each option based on this understanding.

1. Analyze Option (i): The enthalpy of two moles of ZnO is less than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.

   • We know $\Delta H = H_{\text{products}} - H_{\text{reactants}}$.
   • Substituting the given value: $-693.8 \text{ kJ} = H(2 \text{ moles ZnO}) - (H(2 \text{ moles Zn}) + H(1 \text{ mole O}_2))$.
   • Rearranging this equation, we get: $H(2 \text{ moles ZnO}) = (H(2 \text{ moles Zn}) + H(1 \text{ mole O}_2)) - 693.8 \text{ kJ}$.
   • This equation clearly shows that the enthalpy of 2 moles of ZnO (products) is $693.8 \text{ kJ}$ less than the total enthalpy of 2 moles of Zn and 1 mole of O$_2$ (reactants).
   • Therefore, option (i) is correct.

2. Analyze Option (ii): The enthalpy of two moles of ZnO is more than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ. …