Question

Given, \(\mathrm{NH}_{3}(\mathrm{~g})+3 \mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NCl}_{3}(\mathrm{~g})+3 \mathrm{HCl}(\mathrm{g})\) \(-\Delta \mathrm{H}_{1}\) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) ;-\Delta \mathrm{H}_{2}\) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HCl}(\mathrm{g}) ;-\Delta \mathrm{H}_{3}\) The heat of formation of \(\mathrm{NCl}_{3}(\mathrm{~g})\) in terms of \(\Delta \mathrm{H}_{1}, \Delta \mathrm{H}_{2}\) and \(\Delta \mathrm{H}_{3}\) is : (1) \(\Delta \mathrm{H}_{1}=\Delta \mathrm{H}_{1}+\frac{\Delta \mathrm{H}_{2}}{2}-\frac{3}{2} \Delta \mathrm{H}_{3}\) (2) \(\Delta \mathrm{H}_{1}=-\Delta \mathrm{H}_{1}-\frac{\Delta \mathrm{H}_{2}}{2}+\frac{3}{2} \Delta \mathrm{H}_{3}\) (3) \(\Delta \mathrm{H}_{1}=-\Delta \mathrm{H}_{1}+\frac{\Delta \mathrm{H}_{2}}{2}-\frac{3}{2} \Delta \mathrm{H}_{3}\) (4) \(\Delta \mathrm{H}_{\mathrm{t}}=\Delta \mathrm{H}_{1}+\Delta \mathrm{H}_{2}+\Delta \mathrm{H}_{3}\) (5) None of these

Short answer

The correct option is (3): \( \Delta \text{H}_1 = -\Delta \text{H}_1 + \frac{1}{2}\Delta \text{H}_2 - \frac{3}{2}\Delta \text{H}_3 \).

Step-by-step solution

Write the given reactions

Start by writing down all the given reactions: 1. \(\text{NH}_3(\text{g}) + 3 \text{Cl}_2(\text{g}) \rightleftharpoons \text{NCl}_3(\text{g}) + 3 \text{HCl}(\text{g}); -\Delta \text{H}_1\) 2. \(\text{N}_2(\text{g}) + 3 \text{H}_2(\text{g}) \rightleftharpoons 2 \text{NH}_3(\text{g}); -\Delta \text{H}_2\) 3. \(\text{H}_2(\text{g}) + \text{Cl}_2(\text{g}) \rightleftharpoons 2 \text{HCl}(\text{g}); -\Delta \text{H}_3\)

Determine the target reaction

Notice that we need the formation of \(\text{NCl}_3(\text{g})\) from basic elements. The desired reaction can be indirectly derived from the combination of given reactions.

Manipulate reaction 2 for correct stoichiometry

Divide reaction 2 by 2: \(\frac{1}{2}\text{N}_2(\text{g}) + \frac{3}{2}\text{H}_2(\text{g}) \rightleftharpoons \text{NH}_3(\text{g}); -\frac{1}{2}\Delta \text{H}_2\)

Reverse reaction 1

Reverse reaction 1 to form \(\text{NH}_3(\text{g})\) from \(\text{NCl}_3(\text{g})\): \( \text{NCl}_3(\text{g}) + 3 \text{HCl}(\text{g}) \rightleftharpoons \text{NH}_3(\text{g}) + 3 \text{Cl}_2(\text{g}); \Delta \text{H}_1\)

Add the modified and reversed reactions

Combine all steps to get the formation of \(\text{NCl}_3\) from elements: \( \frac{1}{2}\text{N}_2(\text{g}) + \frac{3}{2}\text{H}_2(\text{g}) + 3\text{Cl}_2(\text{g}) \rightleftharpoons \text{NCl}_3(\text{g}) + 3 \text{HCl}(\text{g}) \) with enthalpy changes combined appropriately. Summing the enthalpies, \(\Delta \text{H}_\text{formation} = -\frac{1}{2}\Delta \text{H}_2 + 3(- \frac{1}{2}\Delta \text{H}_3) - \Delta \text{H}_1\)

Simplify the equation

Simplify the result from the last step: \( \Delta \text{H}_\text{formation} = \Delta \text{H}_1 - \frac{1}{2}\Delta \text{H}_2 + (-\frac{3}{2})\Delta \text{H}_3 \)

Key concepts

enthalpy change

Enthalpy change refers to the heat absorbed or released during a chemical reaction at constant pressure. This change is denoted by \(\Delta H\). During exothermic reactions, heat is released, leading to a negative \(\Delta H\). In endothermic reactions, heat is absorbed, resulting in a positive \(\Delta H\).
For instance, in the given exercise:
\[\text{NH}_3(\text{g}) + 3 \text{Cl}_2(\text{g}) \rightleftharpoons \text{NCl}_3(\text{g}) + 3 \text{HCl}(\text{g}) ; -\Delta H_1\] indicates an exothermic reaction, as denoted by the negative sign of \(-\Delta H_1\).
Understanding enthalpy change helps in predicting whether a reaction absorbs or releases energy and the amount involved.

Hess's Law

Hess's Law states that the total enthalpy change for a reaction is the same, regardless of the number of intermediate steps. This is because enthalpy is a state function, depending only on the initial and final states, not on the path taken.
In the exercise provided, Hess’s Law is used to find the heat of formation of \(\text{NCl}_3(\text{g})\) by combining given reactions. The steps involve manipulating and combining reactions ensuring the sums of \(abla H\) values fit the target equation.
Using reaction manipulations:
\[ \text{NH}_3(\text{g}) + 3 \text{Cl}_2(\text{g}) \rightleftharpoons \text{NCl}_3(\text{g}) + 3 \text{HCl}(\text{g}); -\Delta H_1 \] \[ \text{N}_2(\text{g}) + 3 \text{H}_2(\text{g}) \rightleftharpoons 2 \text{NH}_3(\text{g}); -\Delta H_2 \] \[ \text{H}_2(\text{g}) + \text{Cl}_2(\text{g}) \rightleftharpoons 2 \text{HCl}(\text{g}); -\Delta H_3 \] enable the calculation of the formation enthalpy. Students can apply Hess's Law to break down complex reactions into simpler steps and sum their enthalpy changes to find the overall reaction enthalpy.

stoichiometry

Stoichiometry deals with the quantitative relationship between reactants and products in a balanced chemical reaction. It allows for the calculation of the amounts of substances consumed and produced in reactions.
For the given reactions, stoichiometry helps us understand the molar relationships and manipulate the equations accordingly. For example:
\[\text{N}_2(\text{g}) + 3 \text{H}_2(\text{g}) \rightleftharpoons 2 \text{NH}_3(\text{g}); -\Delta H_2\] can be divided by 2 to give:
\[\frac{1}{2} \text{N}_2(\text{g}) + \frac{3}{2} \text{H}_2(\text{g}) \rightleftharpoons \text{NH}_3(\text{g}); -\frac{1}{2} \Delta H_2\] This maintains the stoichiometric ratios to align with other equations in the problem.
Mastering stoichiometry is crucial for balancing equations, understanding reaction yields, and approaching complex chemical problems efficiently. It ensures that the principles of mass conservation are upheld in any chemical reaction.