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Using the following data, calculate the standard heat of formation of \(\operatorname{ICl}(g)\) in \(\mathrm{kJ} / \mathrm{mol}\) : $$ \begin{aligned} \mathrm{Cl}_{2}(g) & \longrightarrow 2 \mathrm{Cl}(g) & \Delta H^{\circ} &=242.3 \mathrm{~kJ} \\ \mathrm{I}_{2}(g) & \longrightarrow 2 \mathrm{I}(g) & \Delta H^{\circ} &=151.0 \mathrm{~kJ} \\ \mathrm{ICl}(g) & \longrightarrow \mathrm{I}(g)+\mathrm{Cl}(g) & \Delta H^{\circ} &=211.3 \mathrm{~kJ} \\ \mathrm{I}_{2}(s) & \Delta H^{\circ}=62.8 \mathrm{~kJ} \end{aligned} $$

Short Answer

Expert verified
The standard heat of formation of ICl(g) is \(-14.65 \mathrm{~kJ/mol}\).

Step by step solution

01

Identify the direct reaction pathway

First, we have to identify the direct reaction pathway from reactants to products for ICl(g). We can see that we need 1/2 I2(g), 1/2 Cl2(g), and ICl(g) reactions.
02

Determine the stoichiometric coefficients

In this step, we will determine the stoichiometric coefficients for our direct reaction pathway. The balanced equation for the formation of ICl(g) is: $$ \frac{1}{2} \mathrm{I}_{2}(g) + \frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{ICl}(g) $$
03

Calculate the standard heat of formation

To calculate the standard heat of formation, we will add or subtract the \(\Delta H^{\circ}\) of each reaction in our direct pathway, making sure to multiply each \(\Delta H^{\circ}\) by its respective stoichiometric coefficient from Step 2. The equation we need to follow is: $$ \Delta H_{f}^{\circ}(\mathrm{ICl}(g)) = \frac{1}{2}\Delta H^{\circ}(\mathrm{I}_{2}(g)) + \frac{1}{2}\Delta H^{\circ}(\mathrm{Cl}_{2}(g)) - \Delta H^{\circ}(\mathrm{ICl}(g)) $$ Using the provided data: $$ \Delta H_{f}^{\circ}(\mathrm{ICl}(g)) = \frac{1}{2}(151.0 \mathrm{~kJ}) + \frac{1}{2}(242.3 \mathrm{~kJ}) - 211.3\mathrm{~kJ} $$
04

Calculate the final answer

Evaluate the expression to find the standard heat of formation for ICl(g): $$ \Delta H_{f}^{\circ}(\mathrm{ICl}(g)) = 75.5\mathrm{~kJ} + 121.15\mathrm{~kJ} - 211.3\mathrm{~kJ} = -14.65\mathrm{~kJ} $$ Thus, the standard heat of formation of ICl(g) is -14.65 kJ/mol.

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Key Concepts

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

Stoichiometric Coefficients
Stoichiometric coefficients play a pivotal role in chemical reactions, especially when it comes to calculating thermodynamic quantities like the standard heat of formation. These coefficients indicate the proportion of each reactant and product involved in a reaction. For any balanced chemical equation, stoichiometric coefficients ensure that the law of conservation of mass is upheld and the number of atoms for each element is the same on both sides of the equation.
  • In this exercise, the stoichiometric coefficients for iodine (\(I_2 ext{(g)}\)) and chlorine (\(Cl_2 ext{(g)}\)) were determined as \(\frac{1}{2}\), meaning that half a mole of each gas is needed to form one mole of iodine monochloride (\(ICl ext{(g)}\)).
  • These coefficients help in scaling the enthalpy change (\(\Delta H\)) for the reaction, appropriately reflecting the change per mole of \(ICl ext{(g)}\).
Understanding stoichiometric coefficients allows us to accurately use given reaction data in thermochemical calculations, which is essential for predicting the enthalpy changes in chemical reactions.
Direct Reaction Pathway
The concept of a direct reaction pathway simplifies the process of calculating thermodynamic changes by outlining a straightforward path from reactants to the desired products.
It essentially allows chemists to break down complex reactions into a series of simpler steps for which data is available.
  • In this context, creating \(ICl ext{(g)}\) involves determining a direct pathway from the elements \(I_2 ext{(g)}\) and \(Cl_2 ext{(g)}\) to the product.
  • Instead of dealing directly with solid iodine, understanding the gaseous pathway helps us use more commonly available \(\Delta H^\circ\) values of intermediate steps.
By using a direct reaction pathway, the task of compiling accurate thermochemical calculations becomes much more manageable, as it allows for the use of available thermodynamic data.
Thermochemistry Calculations
Thermochemistry calculations are crucial for understanding the heat changes that accompany chemical reactions. In this exercise, calculating the standard heat of formation involves combining the \(\Delta H^\circ\) values from various reactions using stoichiometric coefficients.

This is achieved with the equation:\[\Delta H_{f}^{\circ}(ICl(g)) = \frac{1}{2}\Delta H^{\circ}(I_2(g)) + \frac{1}{2}\Delta H^{\circ}(Cl_2(g)) - \Delta H^{\circ}(ICl(g))\]
  • Each term in this equation represents an enthalpy change that associates with transforming the reactants into products along the identified pathway.
  • Using the values from the given data, we substitute them into this equation and find the value of \(\Delta H_{f}^{\circ}(ICl(g))\), which simplifies into evaluating simple arithmetic calculations.
The resulting value of \(-14.65\text{kJ/mol}\) reflects the energy change when one mole of \(ICl\) is formed, giving a comprehensive understanding of the thermodynamic aspects of the reaction. By mastering these calculations, students are able to predict not just energy changes but also aspects like reaction feasibility and direction in thermochemical contexts.

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Most popular questions from this chapter

When \(1.00 \mathrm{~L}\) of \(2.00 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution at \(30.0^{\circ} \mathrm{C}\) is added to \(2.00 \mathrm{~L}\) of \(0.750 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) solution at \(30.0^{\circ} \mathrm{C}\) in a calorimeter, a white solid \(\left(\mathrm{BaSO}_{4}\right)\) forms. The temperature of the mixture increases to \(42.0^{\circ} \mathrm{C}\). Assuming that the specific heat capacity of the solution is \(6.37 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and that the density of the final solution is \(2.00 \mathrm{~g} / \mathrm{mL}\), calculate the enthalpy change per mole of \(\mathrm{BaSO}_{4}\) formed.

Consider \(5.5 \mathrm{~L}\) of a gas at a pressure of \(3.0 \mathrm{~atm}\) in a cylinder with a movable piston. The external pressure is changed so that the volume changes to \(10.5 \mathrm{~L}\). a. Calculate the work done, and indicate the correct sign. b. Use the preceding data but consider the process to occur in two steps. At the end of the first step, the volume is \(7.0 \mathrm{~L}\). The second step results in a final volume of \(10.5\) L. Calculate the work done, and indicate the correct sign. c. Calculate the work done if after the first step the volume is \(8.0 \mathrm{~L}\) and the second step leads to a volume of \(10.5 \mathrm{~L}\). Does the work differ from that in part b? Explain.

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?

Consider the following equations: $$ \begin{aligned} 3 \mathrm{~A}+6 \mathrm{~B} \longrightarrow & 3 \mathrm{D} & \Delta H &=-403 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{E}+2 \mathrm{~F} & \longrightarrow \mathrm{A} & \Delta H &=-105.2 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{C} & \longrightarrow \mathrm{E}+3 \mathrm{D} & \Delta H &=+64.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Suppose the first equation is reversed and multiplied by \(\frac{1}{6}\), the second and third equations are divided by 2, and the three adjusted equations are added. What is the net reaction and what is the overall heat of this reaction?

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