Consider the reaction pathways shown below - Scottish Highers Chemistry - Question 17 - 2018

To determine the enthalpy change for reaction X, we apply Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for individual steps.

We have the following reactions:

1. Formation of CO from C and O₂: $C(s) + O_2(g) → CO(g) + \frac{1}{2}O_2(g) \quad (\Delta H = -394\, kJ\, mol^{-1})$
2. Formation of CO₂ from CO and O₂: $CO(g) + \frac{1}{2}O_2(g) → CO_2(g) \quad (\Delta H = -283\, kJ\, mol^{-1})$

Now, we need to write the equation for reaction X and combine the enthalpy changes:

The enthalpy change for reaction X can be found as follows:

$\Delta H_X = \Delta H_{CO} + \Delta H_{O_2}$ $\Delta H_X = -394 + (-283)$ $\Delta H_X = -394 - 283 = -677\, kJ\, mol^{-1}$

However, we need the enthalpy change for the formation of X in the reverse direction to yield a positive value. Thus:

$\Delta H = +677\, kJ\, mol^{-1}$

So, the answer corresponds to the option B: +111 kJ mol⁻¹, which is the required value for reaction X.