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Use the following standard enthalpies of formation to estimate the \(\mathrm{N}-\mathrm{H}\) bond energy in ammonia: \(\mathrm{N}(g), 472.7 \mathrm{kJ} / \mathrm{mol} ; \mathrm{H}(g)\) \(216.0 \mathrm{kJ} / \mathrm{mol} ; \mathrm{NH}_{3}(g),-46.1 \mathrm{kJ} / \mathrm{mol} .\) Compare your value to the one in Table \(3-3\).

Short Answer

Expert verified
The N-H bond energy in ammonia can be calculated using the given enthalpies of formation: ∆H = -46.1 - (\( \dfrac{1}{2}\) x 472.7 + \( \dfrac{3}{2}\) x 216.0) = -393.0 kJ/mol. Dividing this by the three N-H bonds present in ammonia, we find the N-H bond energy to be -131.0 kJ/mol. Comparing this value with the one given in Table 3-3 allows us to assess the accuracy and reliability of the data and calculations.

Step by step solution

01

Calculate the enthalpy change for the formation of one mole of ammonia

To calculate the enthalpy change for the formation of one mole of ammonia, we need to consider the following reaction: \( \dfrac{1}{2}\ N_{2}(g) + \dfrac{3}{2}\ H_{2}(g) \rightarrow NH_{3}(g) \) The enthalpy change for this reaction, ∆H, can be calculated using the standard enthalpies of formation for each species involved in the reaction, according to the formula: ∆H = ∆Hf(products) - ∆Hf(reactants) We are given the enthalpy of formation for each species: ∆Hf(N(g))= 472.7 kJ/mol, ∆Hf(H(g))= 216.0 kJ/mol, ∆Hf(NH₃(g))= -46.1 kJ/mol. Plugging these values into the formula, we get: ∆H = (-46.1) - (\( \dfrac{1}{2}\) x 472.7 + \( \dfrac{3}{2}\) x 216.0) = -393.0 kJ/mol
02

Calculate the N-H bond energy

Since there are three N-H bonds in one mole of ammonia, we can calculate the N-H bond energy by dividing the enthalpy change by the number of bonds: N-H bond energy = ∆H / 3 = (-393.0 kJ/mol) / 3 = -131.0 kJ/mol
03

Compare the calculated N-H bond energy with the value given in Table 3-3

Now that we have calculated the N-H bond energy in ammonia, we can compare it with the value given in Table 3-3. If there is a slight difference between the two values, it may be due to various factors such as experimental errors, rounding, or simplifications in calculations. If the values are very similar, this would suggest that our calculations are accurate and that the given data is reliable.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy Change
Enthalpy change is an essential concept in chemistry, particularly in thermodynamics. It represents the heat change that occurs in a system at a constant pressure during a chemical reaction. This can either be an exothermic reaction, where heat is released, or an endothermic reaction, where heat is absorbed.
When dealing with reactions, such as the formation of ammonia, we often use the equation:
  • \( \Delta H = \Delta H_f(\text{products}) - \Delta H_f(\text{reactants}) \)
This formula helps us calculate the enthalpy change by using the standard enthalpies of formation. For example, in forming ammonia, we calculate the enthalpy change by subtracting the sum of the enthalpies of the gases involved from the enthalpy of ammonia. Understanding enthalpy change is crucial for predicting the energy dynamics in chemical reactions.
Enthalpies of Formation
Enthalpies of formation refer to the heat change when one mole of a compound is formed from its elements in their standard states. These values are crucial in calculating the overall enthalpy change in chemical reactions.
For ammonia, we have:
  • \(\Delta H_f (\text{N}_2) \)
  • \(\Delta H_f (\text{H}_2) \)
  • \(\Delta H_f (\text{NH}_3) \)
These values are typically tabulated, allowing for straightforward calculations in computing reaction enthalpies. Standard enthalpies of formation provide insight into the stability of compounds and are fundamental for various calculations in chemistry, particularly involving predictive and comparative analyses of energy changes.
Ammonia
Ammonia, chemically represented as \(\text{NH}_3\), plays a vital role in the chemical industry and nature. It is a compound of nitrogen and hydrogen, and it is known for its pungent smell and ability to react readily with other substances.
The formation of ammonia can be represented by the reaction:
  • \( \frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \rightarrow \text{NH}_3(g) \)
In this reaction, three hydrogen atoms chemically bond with one nitrogen atom, resulting in an N-H bond. Studying the energy associated with these bonds is crucial, as it informs us about the stability and reactivity of the compound.
Ammonia is not only significant in agricultural fertilizers but also in many industrial processes where its formation and energy dynamics are key considerations.

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Most popular questions from this chapter

For the process \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\) at \(298 \mathrm{K}\) and 1.0 atm, \(\Delta H\) is more positive than \(\Delta E\) by 2.5 kJ/mol. What does the \(2.5 \mathrm{kJ} / \mathrm{mol}\) quantity represent?

In a coffee-cup calorimeter, \(50.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{AgNO}_{3}\) and \(50.0 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HCl are mixed to yield the following reaction: $$\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s)$$ The two solutions were initially at \(22.60^{\circ} \mathrm{C},\) and the final temperature is \(23.40^{\circ} \mathrm{C}\). Calculate the heat that accompanies this reaction in \(\mathrm{kJ} / \mathrm{mol}\) of \(\mathrm{AgCl}\) formed. Assume that the combined solution has a mass of \(100.0 \mathrm{g}\) and a specific heat capacity of \(4.18 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\).

Consider the following reaction: $$\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-891 \mathrm{kJ}$$ Calculate the enthalpy change for each of the following cases: a. \(1.00 \mathrm{g}\) methane is burned in excess oxygen. b. \(1.00 \times 10^{3} \mathrm{L}\) methane gas at \(740 .\) torr and \(25^{\circ} \mathrm{C}\) are burned in excess oxygen. The density of \(\mathrm{CH}_{4}(g)\) at these conditions is \(0.639 \mathrm{g} / \mathrm{L}\).

The standard enthalpy of combustion of ethene gas, \(\mathrm{C}_{2} \mathrm{H}_{4}(g)\) is \(-1411.1 \mathrm{kJ} / \mathrm{mol}\) at \(298 \mathrm{K}\). Given the following enthalpies of formation, calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{C}_{2} \mathrm{H}_{4}(g)\). $$\begin{array}{ll}\mathrm{CO}_{2}(g) & -393.5 \mathrm{kJ} / \mathrm{mol} \\\\\mathrm{H}_{2} \mathrm{O}(l) & -285.8 \mathrm{kJ} / \mathrm{mol}\end{array}$$

What is meant by the term lower in energy? Which is lower in energy, a mixture of hydrogen and oxygen gases or liquid water? How do you know? Which of the two is more stable? How do you know?

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