Q.On the basis of thermochemical equations (a),

The key idea here is Hess's Law of Constant Heat Summation.

Hess's Law states that the total enthalpy change for a chemical reaction is the same, regardless of the number of steps taken to complete the reaction, provided the initial and final states are identical.

1. We are given three thermochemical equations:

   1. C (graphite) + O$_2$(g) → CO$_2$(g) ; $\Delta_rH = x$
   2. C (graphite) + (1/2) O$_2$(g) → CO(g) ; $\Delta_rH = y$
   3. CO(g) + (1/2) O$_2$(g) → CO$_2$(g) ; $\Delta_rH = z$

2. Observe that if we add equations (b) and (c), the intermediate product CO(g) cancels out:

   [C (graphite) + (1/2) O$_2$(g) → CO(g)] + [CO(g) + (1/2) O$_2$(g) → CO$_2$(g)]

   C (graphite) + (1/2) O$_2$(g) + CO(g) + (1/2) O$_2$(g) → CO(g) + CO$_2$(g) …

The problem asks us to find the correct algebraic relationship between the enthalpy changes ($x$, $y$, and $z$) of three given thermochemical equations. This type of problem is a direct application of Hess's Law of Constant Heat Summation.

Concept and Intuition: Hess's Law

Enthalpy ($H$) is a state function, meaning its value depends only on the initial and final states of a system, not on the path taken to get there. The change in enthalpy ($\Delta H$) for a reaction, therefore, is also independent of the reaction pathway.

Imagine you want to travel from City A to City C. You could take a direct flight (Path 1), or you could fly from City A to City B, and then from City B to City C (Path 2). The total displacement (change in position) from A to C is the same regardless of whether you took the direct flight or the two-leg journey.

Similarly, in chemistry, if a chemical reaction can occur in one step or in a series of steps, the total enthalpy change for the overall reaction will be the same whether it occurs in one step or multiple steps. This allows us to calculate the enthalpy change of a reaction by algebraically combining the enthalpy changes of other known reactions.

Step-by-step Solution

We are given three thermochemical equations:

1. C (graphite) + O$_2$(g) → CO$_2$(g) ; $\Delta_r H = x$ kJ mol$^{-1}$
2. C (graphite) + (1/2) O$_2$(g) → CO(g) ; $\Delta_r H = y$ kJ mol$^{-1}$
3. CO(g) + (1/2) O$_2$(g) → CO$_2$(g) ; $\Delta_r H = z$ kJ mol$^{-1}$ Our goal is to find a relationship between $x$, $y$, and $z$. We can achieve this by trying to combine reactions (b) and (c) to see if they yield reaction (a), or vice versa.

1. Analyze the overall transformation:

   Reaction (a) represents the complete combustion of graphite to carbon dioxide.

   C (graphite) $\rightarrow$ CO$_2$(g)

2. Consider the intermediate steps:

   Reactions (b) and (c) represent a two-step process for the same overall transformation:

   Step 1 (reaction b): C (graphite) is first oxidized to carbon monoxide.

   C (graphite) $\rightarrow$ CO(g)

   Step 2 (reaction c): The carbon monoxide is then further oxidized to carbon dioxide.

   CO(g) $\rightarrow$ CO$_2$(g)

3. Combine reactions (b) and (c): …