Question

The wrong statement among the following is (1) The heat change for the reaction \(\mathrm{II}_{2} \mathrm{O}(1) \rightarrow\) \(\mathrm{II}_{2} \mathrm{O}(\mathrm{g})\) is known as heat of vapourisation. (2) The heat change in the reaction \(\mathrm{C}(\mathrm{s})+2 \mathrm{~S}(\mathrm{~s}) \rightarrow\) \(\mathrm{CS}_{2}\) (1) is called heat of formation of \(\mathrm{CS}_{2}\). (3) The standard heat cnthalpy of diamond is zero. (4) The enthalpy change \(\mathrm{C}_{(\mathrm{s})} \rightarrow \mathrm{C}_{(\mathrm{g})}\) is known as cnthalpy of sublimation.

Short answer

Statement 3 is incorrect. The standard enthalpy of formation for diamond is not zero.

Step-by-step solution

Identify the statements

Read all the given statements carefully and identify each one by recognizing the type of heat change they are referring to.

Evaluate statement 1

Evaluate the first statement: 'The heat change for the reaction \( H_{2}O(\text{l}) \rightarrow H_{2}O(\text{g}) \) is known as heat of vaporization'. This is correct, as the heat change from liquid to gas is indeed called the heat of vaporization.

Evaluate statement 2

Evaluate the second statement: 'The heat change in the reaction \( \text{C}(\text{s}) + 2\text{S}(\text{s}) \rightarrow \text{CS}_{2}(\text{l}) \) is called heat of formation of \( \text{CS}_{2} \)'. This is correct, as the heat of formation is the enthalpy change when one mole of a compound is formed from its constituent elements under standard conditions.

Evaluate statement 3

Evaluate the third statement: 'The standard heat enthalpy of diamond is zero'. This is incorrect because the standard enthalpy of formation for diamond is not zero. The standard enthalpy of formation for graphite is zero, not diamond.

Evaluate statement 4

Evaluate the fourth statement: 'The enthalpy change \( C_{(\text{s})} \rightarrow C_{(\text{g})} \) is known as enthalpy of sublimation'. This is correct, as the enthalpy of sublimation refers to the phase change from solid to gas.

Key concepts

heat of vaporization

The heat of vaporization is the amount of energy required to convert a liquid into a gas at its boiling point. This process involves breaking the intermolecular forces that hold the liquid molecules together. For example, when water (H₂O) is heated and changes from liquid to steam, it requires a specific amount of heat known as the heat of vaporization. The formula to calculate this is given by:
\[ q = m \times \text{Δ}H_v \]
Where:

• q is the heat added
• m is the mass
• ΔH_v is the heat of vaporization

heat of formation

The heat of formation is the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states. It is also known as standard enthalpy of formation. For example, the reaction for the formation of carbon disulfide (CS₂) from carbon and sulfur is:
C(s) + 2S(s) → CS₂(l)
Calculating the heat of formation involves adding the enthalpy changes for each step of the reaction. This value is essential for predicting the stability of a compound and its heat content compared to its elements.

enthalpy of sublimation

The enthalpy of sublimation is the heat required to convert a solid directly into a gas without passing through the liquid state. This involves overcoming the attractions between the particles in the solid. The process is endothermic, meaning it requires the input of energy. For carbon, this can be represented as:
C(s) → C(g)
The enthalpy of sublimation can be found using Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step. The value is unique for each substance and depends on its molecular structure.

standard enthalpy

Standard enthalpy, often referred to as the standard enthalpy change, is the heat change associated with a reaction that occurs under standard conditions (1 atm pressure and 298 K temperature). This provides a baseline for comparing enthalpy changes of different reactions. For example, the standard enthalpy of formation for graphite (a form of carbon) is zero because it is in its most stable form. However, this is not true for diamond, which has a different structure and higher enthalpy. This is why the standard enthalpy of formation is crucial in identifying the stability and energy content of different forms of elements or compounds.