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Chapter 6: Problem 114

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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Most popular questions from this chapter

It has been determined that the body can generate \(5500 \mathrm{~kJ}\) of energy during one hour of strenuous exercise. Perspiration is the body's mechanism for eliminating this heat. What mass of water would have to be evaporated through perspiration to rid the body of the heat generated during two hours of exercise? (The heat of vaporization of water is \(40.6 \mathrm{~kJ} / \mathrm{mol}\).)

The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ a. Use the values of \(\Delta H_{\mathrm{f}}^{\circ}\) in Appendix 4 to calculate the value of \(\Delta H^{\circ}\) for each of the preceding reactions. b. Write the overall equation for the production of nitric acid by the Ostwald process by combining the preceding equations. (Water is also a product.) Is the overall reaction exothermic or endothermic?

Combustion of table sugar produces \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) When \(1.46 \mathrm{~g}\) table sugar is combusted in a constant-volume (bomb) calorimeter, \(24.00 \mathrm{~kJ}\) of heat is liberated. a. Assuming that table sugar is pure sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)\), write the balanced equation for the combustion reaction. b. Calculate \(\Delta E\) in \(\mathrm{kJ} / \mathrm{mol} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) for the combustion reaction of sucrose. c. Calculate \(\Delta I I\) in \(\mathrm{kJ} / \mathrm{mol} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) for the combustion reaction of sucrose at \(25^{\circ} \mathrm{C}\).

Consider the dissolution of \(\mathrm{CaCl}_{2}\) : $$ \mathrm{CaCl}_{2}(s) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad \Delta H=-81.5 \mathrm{~kJ} $$ An \(11.0-\mathrm{g}\) sample of \(\mathrm{CaCl}_{2}\) is dissolved in \(125 \mathrm{~g}\) water, with both substances at \(25.0^{\circ} \mathrm{C}\). Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\).

Given the following data $$ \begin{aligned} \mathrm{Ca}(s)+2 \mathrm{C}(\text { graphite }) & \longrightarrow \mathrm{CaC}_{2}(s) & \Delta H &=-62.8 \mathrm{~kJ} \\ \mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CaO}(s) & \Delta H &=-635.5 \mathrm{~kJ} \\ \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q) & \Delta H &=-653.1 \mathrm{~kJ} \\ \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-1300 . \mathrm{kJ} \\ \mathrm{C}(\text { graphite })+\mathrm{O}_{2}(g) & \Delta H &=-393.5 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g) $$
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