Chapter 9: Problem 58
Could an endothermic reaction be used to determine the heat capacity of a calorimeter?
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Chapter 9: Problem 58
Could an endothermic reaction be used to determine the heat capacity of a calorimeter?
These are the key concepts you need to understand to accurately answer the question.
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Breaking the small pouch of water inside a chemical cold pack containing ammonium nitrate activates the pack, which is used by sports trainers for injured athletes. What is the sign of \(\Delta H\) for the process taking place in the cold pack?
The destruction of the ozone layer by chlorofluorocarbons (CFCs) can be described by the following reactions: $$\begin{aligned}\mathrm{ClO}(g)+\mathrm{O}_{3}(g) & \rightarrow \mathrm{Cl}(g)+2 \mathrm{O}_{2}(g) & \Delta H_{\mathrm{rxn}}^{\circ} &=-29.90 \mathrm{kJ} \\\2 \mathrm{O}_{3}(g) \rightarrow 3 \mathrm{O}_{2}(g) & & \Delta H_{\mathrm{rxn}}^{\circ} &=+24.18 \mathrm{kJ}\end{aligned}$$ Use the preceding \(\Delta H_{\mathrm{rxn}}^{\circ}\) values to determine the value of the standard heat of reaction for this reaction: $$\mathrm{Cl}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=?$$
The mineral spodumene \(\left(\operatorname{LiAlSi}_{2} \mathrm{O}_{6}\right)\) exists in two crystalline forms called \(\alpha\) and \(\beta .\) Use Hess's law and the following information to calculate \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the conversion of \(\alpha\) -spodumene into \(\beta\) -spodumene: $$\begin{array}{r}\mathrm{Li}_{2} \mathrm{O}(s)+2 \mathrm{Al}(s)+4 \mathrm{SiO}_{2}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow 2 \alpha-\mathrm{LiAlSi}_{2} \mathrm{O}_{6}(s) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-1870.6 \mathrm{kJ} \end{array}$$ $$\begin{array}{r}\mathrm{Li}_{2} \mathrm{O}(s)+2 \mathrm{Al}(s)+4 \mathrm{SiO}_{2}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow 2 \beta-\mathrm{LiAlSi}_{2} \mathrm{O}_{6}(s) \\\\\Delta H_{\mathrm{rxn}}^{\circ}=-1814.6 \mathrm{kJ} \end{array}$$
How is Hess's law consistent with the law of conservation of energy?
In a simple "kitchen chemistry" experiment, some vinegar is poured into an empty soda bottle. A deflated balloon containing baking soda is stretched over the mouth of the bottle. Holding up the balloon and shaking it allows the baking soda to fall into the vinegar, which starts the following reaction and inflates the balloon: $$\begin{aligned} \mathrm{NaHCO}_{3}(a q)+\mathrm{CH}_{3} \mathrm{COOH}(a q) \rightarrow & \\ & \mathrm{CH}_{3} \mathrm{COONa}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell) \end{aligned}$$ If the contents of the bottle are the system, is work being done on the surroundings or on the system?
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