Chapter 10: Problem 14
State Hess's law. How is it used?
Short Answer
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 10: Problem 14
State Hess's law. How is it used?
These are the key concepts you need to understand to accurately answer the question.
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Get started for freeCalculate the enthalpy of formation for sulfur dioxide, \(\mathrm{SO}_{2},\) from its elements, sulfur and oxygen. Use the balanced chemical equation and the following information. \begin{equation} \begin{array}{c}{\mathrm{S}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{3}(g) \quad \Delta H=-395.8 \mathrm{kJ} / \mathrm{mol}} \\ {2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g) \quad \Delta H=-198.2 \mathrm{kJ} / \mathrm{mol}}\end{array} \end{equation}
Graphing Calculator Calculating the Gibbs-Energy Change The graphing calculator can run a program that calculates the Gibbs-energy change, given the temperature, \(T\) , change in enthalpy \(\Delta H,\) and change in entropy, \(\Delta S .\) Given that the temperature is 298 \(\mathrm{K}\) , the change in enthalpy is 131.3 \(\mathrm{kJ} / \mathrm{mol}\) , and the change in entropy is \(0.134 \mathrm{kJ} /(\mathrm{mol} \cdot \mathrm{K}),\) you can calculate Gibbs-energy change in kilojoules per mole. Then use the program to make calculations. Go to Appendix C. If you are using a TI-83 Plus, you can download the program ENERGY data and run the application as directed. If you are using another calculator, your teacher will provide you with keystrokes and data sets to use. After you have run the program, answer the following questions. \begin{equation} \begin{array}{l}{\text { a. What is the Gibbs-energy change given a }} \\\ {\text { temperature of } 300 \mathrm{K}, \text { a change in enthalpy }} \\\ {\text { of } 132 \mathrm{kJ} / \mathrm{mol} \text { and a change in entropy of }} \\ {0.086 \mathrm{kJ} / /(\mathrm{mol} \cdot \mathrm{K}) ?}\\\\{\text { b. What is the Gibbs-energy change given a }} \\ {\text { temperature of } 288 \mathrm{K}, \text { a change in enthalpy }} \\ {\text { of } 115 \mathrm{kJ} / \mathrm{mol} \text { and a change in entropy of }} \\ {0.113 \mathrm{kJ} /(\mathrm{mol} \cdot \mathrm{K}) ?}\\\\{\text { c. What is the Gibbs-energy change given a }} \\ {\text { temperature of } 298 \mathrm{K} \text { , a change in enthalpy }} \\ {\text { of } 181 \mathrm{kJ} / \mathrm{mol} \text { and a change in entropy of }} \\ {0.135 \mathrm{kJ} /(\mathrm{mol} \cdot \mathrm{K}) ?}\end{array} \end{equation}
Distinguish between heat and temperature.
You need 70.2 \(\mathrm{J}\) to raise the temperature of 34.0 \(\mathrm{g}\) of ammonia, \(\mathrm{NH}_{3}(g),\) from \(23.0^{\circ} \mathrm{C}\) to \(24.0^{\circ} \mathrm{C}\) . Calculate the molar heat capacity of ammonia.
At high temperatures, does enthalpy or entropy have a greater effect on a reaction's Gibbs energy?
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