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Chapter 8: Problem 48

When one mole of calcium carbonate reacts with ammonia, solid calcium cyanamide, \(\mathrm{CaCN}_{2}\), and liquid water are formed. The reaction absorbs \(90.1 \mathrm{~kJ}\) of heat. (a) Write a balanced thermochemical equation for the reaction. (b) Using Table 8.3, calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for calcium cyanamide.

Short Answer

Expert verified
Answer: The standard enthalpy of formation of calcium cyanamide in the reaction between calcium carbonate and ammonia is -619.5 kJ/mol.

Step by step solution

01

(Step 1: Write the balanced chemical equation)

The first step to solve this problem is to write the balanced chemical equation for the reaction between calcium carbonate and ammonia. The reactants are calcium carbonate (CaCO3) and ammonia (NH3), and the products are calcium cyanamide (CaCN2) and water (H2O) in its liquid state. CaCO3 (s) + 2NH3 (g) -> CaCN2 (s) + CO2 (g) + H2O (l)
02

(Step 2: Write the thermochemical equation)

The second step is to include the heat absorbed, \(90.1 \mathrm{~kJ}\), in the balanced chemical equation. Since the reaction absorbs heat, it is an endothermic reaction. We can write the enthalpy change as \(\Delta H = +90.1 \mathrm{~kJ}\). CaCO3 (s) + 2NH3 (g) -> CaCN2 (s) + CO2 (g) + H2O (l) + \(90.1 \mathrm{kJ}\)
03

(Step 3: Calculate the enthalpy of formation of calcium cyanamide)

To calculate the standard enthalpy of formation of calcium cyanamide, we need to use the enthalpy of formation of each substance involved in the reaction, using Table 8.3 data. Using Hess's law, we can calculate the enthalpy of formation of calcium cyanamide (\(\Delta H_{\mathrm{f}}^{\circ}\)) as: \(\Delta H_{\mathrm{f}}^{\circ}(\text{CaCN}_2) = \Delta H_{\mathrm{f}}^{\circ}(\text{CaCO}_3) + 2\Delta H_{\mathrm{f}}^{\circ}(\text{NH}_3) - \Delta H_{\mathrm{f}}^{\circ}(\text{CO}_2) - \Delta H_{\mathrm{f}}^{\circ}(\text{H}_2 \text{O})\) We just need to substitute the values from Table 8.3 and solve for \(\Delta H_{\mathrm{f}}^{\circ}(\text{CaCN}_2)\). \(\Delta H_{\mathrm{f}}^{\circ}(\text{CaCO}_3) = -1207.0 \, \mathrm{kJ/mol}\) \(\Delta H_{\mathrm{f}}^{\circ}(\text{NH}_3) = -45.9 \, \mathrm{kJ/mol}\) \(\Delta H_{\mathrm{f}}^{\circ}(\text{CO}_2) = -393.5 \, \mathrm{kJ/mol}\) \(\Delta H_{\mathrm{f}}^{\circ}(\text{H}_2 \text{O}) = -285.8 \, \mathrm{kJ/mol}\) \(\Delta H_{\mathrm{f}}^{\circ}(\text{CaCN}_2) = (-1207.0) + 2(-45.9) - (-393.5) - (-285.8)\)
04

(Step 4: Calculate the final answer)

Now, we can calculate the final answer: \(\Delta H_{\mathrm{f}}^{\circ}(\text{CaCN}_2) = -1207.0 + (-91.8) + 393.5 + 285.8 = -619.5 \, \mathrm{kJ/mol}\) The standard enthalpy of formation of calcium cyanamide is -619.5 kJ/mol.

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Key Concepts

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

Thermochemical Equations
A thermochemical equation is a balanced chemical equation that includes the enthalpy change for the reaction. It tells us not only what reacts and what is formed, but also how much heat is absorbed or released.
For the given reaction, where calcium carbonate reacts with ammonia to form calcium cyanamide, the equation would be:
\[ \text{CaCO}_3 (s) + 2\text{NH}_3 (g) \rightarrow \text{CaCN}_2 (s) + \text{CO}_2 (g) + \text{H}_2\text{O} (l) + 90.1 \text{ kJ} \]
The positive sign of 90.1 kJ indicates that the reaction absorbs heat. This is what makes it an endothermic reaction.
Including the heat change in the equation helps us understand how energy is involved and needed in chemical processes.
Hess's Law
Hess's Law is a powerful principle in chemistry stating that the total enthalpy change for a reaction is the same, no matter how many steps it takes. This means we can calculate the enthalpy change using other known reactions.
To find the enthalpy of formation for calcium cyanamide, we use Hess's Law with known values from other substances:
\[\Delta H_{\mathrm{f}}^{\circ}(\text{CaCN}_2) = \Delta H_{\mathrm{f}}^{\circ}(\text{CaCO}_3) + 2\Delta H_{\mathrm{f}}^{\circ}(\text{NH}_3) - \Delta H_{\mathrm{f}}^{\circ}(\text{CO}_2) - \Delta H_{\mathrm{f}}^{\circ}(\text{H}_2 \text{O})\]
By substituting the standard enthalpy values, we find:
\[ \Delta H_{\mathrm{f}}^{\circ}(\text{CaCN}_2) = -619.5 \, \text{kJ/mol}\]
This illustrates how Hess's Law allows us to deduce unknown values using the combined energies of known reactions.
Endothermic Reactions
An endothermic reaction is one where the system absorbs heat from the surroundings. In our reaction, heat intake of 90.1 kJ signifies it's endothermic. These reactions require an input of energy to proceed.
Examples of endothermic reactions include:
  • Melting ice into water
  • Photosynthesis in plants
  • Evaporating water

Understanding whether a reaction is endothermic or exothermic helps us design processes, predict temperature changes, and measure energy efficiency in reactions.
Knowing this about our calcium carbonate reaction aids in anticipating energy needs and yields.

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Most popular questions from this chapter

Consider the following reaction in a vessel with a movable piston. $$ \mathrm{X}(g)+\mathrm{Y}(g) \longrightarrow \mathrm{Z}(l) $$ As the reaction occurs, the system loses \(1185 \mathrm{~J}\) of heat. The piston moves down and the surroundings do \(623 \mathrm{~J}\) of work on the system. What is \(\Delta E ?\)

Given $$ 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) \longrightarrow 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H^{\circ}=3351.4 \mathrm{~kJ} $$ (a) What is the heat of formation of aluminum oxide? (b) What is \(\Delta H^{\circ}\) for the formation of \(12.50 \mathrm{~g}\) of aluminum oxide?

Write thermochemical equations for the decomposition of one mole of the following compounds into the elements in their stable states at \(25^{\circ} \mathrm{C}\) and 1 atm. (a) ethyl alcohol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) (b) sodium fluoride \((s)\) (c) magnesium sulfate \((s)\) (d) ammonium nitrate (s)

Given the following thermochemical equations, $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow & 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & & \Delta H=-1299.5 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & & \Delta H=-393.5 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & & \Delta H=-285.8 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the decomposition of one mole of acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\), to its elements in their stable state at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

To produce silicon, used in semiconductors, from sand \(\left(\mathrm{SiO}_{2}\right)\), a reaction is used that can be broken down into three steps: $$ \begin{aligned} \mathrm{SiO}_{2}(s)+2 \mathrm{C}(s) \longrightarrow \mathrm{Si}(s)+2 \mathrm{CO}(g) & & \Delta H=689.9 \mathrm{~kJ} \\ \mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(g) & & \Delta H=-657.0 \mathrm{~kJ} \\ \mathrm{SiCl}_{4}(g)+2 \mathrm{Mg}(s) \longrightarrow 2 \mathrm{MgCl}_{2}(s)+\mathrm{Si}(s) & & \Delta H=-625.6 \mathrm{~kJ} \end{aligned} $$ (a) Write the thermochemical equation for the overall reaction for the formation of silicon from silicon dioxide; \(\mathrm{CO}\) and \(\mathrm{MgCl}_{2}\) are byproducts. (b) What is \(\Delta H\) for the formation of one mole of silicon? (c) Is the overall reaction exothermic?
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