Question

Calculate the standard enthalpy change, \(\Delta \mathrm{H}^{\circ}\), for the combustion of ammonia, \(\mathrm{NH}_{3}(\mathrm{~g})\), to give nitric oxide, \(\mathrm{NO}(\mathrm{g})\), and water \(\mathrm{H}_{2} \mathrm{O}(\ell)\). The enthalpies of formation, \(\Delta \mathrm{H}^{\circ} \mathrm{f}\), are \(68.32 \mathrm{Kcal} /\) mole for \(\mathrm{H}_{2} \mathrm{O}(\ell),-11.02 \mathrm{Kca} 1 /\) mole for \(\mathrm{NH}_{3}(\mathrm{~g})\), and \(21.57 \mathrm{Kcal} /\) mole for \(\mathrm{NO}(\mathrm{g})\).

Short answer

The standard enthalpy change, ∆H°, for the combustion of ammonia to give nitric oxide and liquid water is 540.28 Kcal, calculated using the formula ∆H°(reaction) = ∑ [∆H°f(products)] - ∑ [∆H°f(reactants)] and the given enthalpies of formation.

Step-by-step solution

Write the balanced chemical equation for the combustion of ammonia

For the combustion of ammonia (NH3) to give nitric oxide (NO) and liquid water (H2O), the balanced chemical equation is: 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l) Now that we have the balanced equation, we can proceed to calculate the standard enthalpy change.

Use the given enthalpies of formation to find the standard enthalpy change

We will apply the formula for the standard enthalpy change: ∆H°(reaction) = ∑ [∆H°f(products)] - ∑ [∆H°f(reactants)] Enthalpies of formation for the compounds are given as follows: - ∆H°f(NH3(g)) = -11.02 Kcal/mol - ∆H°f(NO(g)) = 21.57 Kcal/mol - ∆H°f(H2O(l)) = 68.32 Kcal/mol The enthalpy of formation of O2(g) is 0, as O2 is an element in its standard state. Now, calculate the standard enthalpy change using the balanced equation: ∆H°(reaction) = [4 × ∆H°f(NO) + 6 × ∆H°f(H2O)] - [4 × ∆H°f(NH3) + 5 × ∆H°f(O2)] ∆H°(reaction) = [4 × 21.57 Kcal/mol + 6 × 68.32 Kcal/mol] - [4 × (-11.02) Kcal/mol + 5 × 0 Kcal/mol]

Compute the standard enthalpy change and present the result

Now carry out the mathematical operations: ∆H°(reaction) = (86.28 + 409.92 - (-44.08)) ∆H°(reaction) = 540.28 Kcal The standard enthalpy change, ∆H°, for the combustion of ammonia to give nitric oxide and liquid water is 540.28 Kcal.

Key concepts

Enthalpies of Formation

Enthalpies of formation are crucial for understanding how heat energy changes when substances react chemically. These values represent the amount of energy released or absorbed when one mole of a compound is formed from its elements in their standard states. The standard enthalpy of formation is denoted as \( \Delta H^{\circ}_f \).

For instance, in the combustion reaction of ammonia, knowing the enthalpies of formation for each reactant and product helps in calculating the overall energy change. This involves summing up the enthalpies for the products and subtracting the sum for the reactants.

To compute the standard enthalpy change \( \Delta H^{\circ} \) for a reaction, use the equation:

• \( \Delta H^{\circ}(\text{reaction}) = \sum \Delta H^{\circ}_f(\text{products}) - \sum \Delta H^{\circ}_f(\text{reactants}) \)

This calculation helps determine if a reaction is endothermic (absorbs heat) or exothermic (releases heat).

Combustion Reaction

A combustion reaction is a chemical process where a substance reacts with oxygen, releasing energy in the form of heat and light. This type of reaction typically involves hydrocarbons or other organic molecules. In the given exercise, ammonia \((\text{NH}_3)\) undergoes combustion to form nitric oxide \((\text{NO})\) and water \((\text{H}_2\text{O})\).

Combustion reactions are commonly exothermic, meaning they release energy. These reactions are important in various applications, from powering engines to heating homes.

The reaction provided in the exercise:

• \(4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2\text{O}(l)\)

illustrates how ammonia combusts with oxygen, leading to the formation of new products and energy release.

Balanced Chemical Equation

A balanced chemical equation accurately represents the reactants and products in a chemical reaction, ensuring that the number of atoms for each element is the same on both sides of the equation. This balance reflects the conservation of mass.

In the combustion of ammonia, the balanced equation is:

• \(4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2\text{O}(l)\)

Balancing the equation is crucial because it allows for accurate calculations of the reaction's enthalpy change using enthalpies of formation. Each coefficient in the balanced equation corresponds to the number of moles of a substance, guiding the use of these values in calculations.

This ensures precision when applying the formula for enthalpy change, leading to correct energy assessments for the reaction.

Standard State Elements

Standard state refers to the most stable physical form of an element or compound at a defined set of conditions, usually 1 atmosphere of pressure and a specified temperature, commonly 25°C (298 K). These conditions are used as a reference to measure enthalpies of formation.

Elements in their standard state have a standard enthalpy of formation of zero. For example, oxygen gas \((\text{O}_2)\) in its standard state doesn't contribute energy to the enthalpy change when forming compounds.

Understanding standard states ensures consistency in reporting and calculating energy changes. For the reaction involving ammonia, recognition of standard states helps simplify calculations by setting the baseline enthalpy of formation for elemental oxygen to zero, making the computation of \( \Delta H^{\circ} \) more straightforward.