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Using values from Appendix \(\mathrm{C},\) calculate the value of \(\Delta H^{\circ}\) for each of the following reactions: (a) \(\mathrm{CaO}(s)+2 \mathrm{HCl}(g) \longrightarrow \mathrm{CaCl}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(4 \mathrm{FeO}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) (c) \(2 \mathrm{CuO}(s)+\mathrm{NO}(g) \longrightarrow \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{NO}_{2}(g)\) (d) \(4 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Short Answer

Expert verified
The calculated values of 螖H鈦 for the given reactions are as follows: (a) 螖H鈦 = -217.5 kJ/mol (b) 螖H鈦 = -560.4 kJ/mol (c) 螖H鈦 = 86.5 kJ/mol (d) 螖H鈦 = -286.8 kJ/mol

Step by step solution

01

(a) CaO(s) + 2 HCl(g) -> CaCl2(s) + H2O(g)

First, we need to find the 螖H鈦 values of the reactants and products from Appendix C. 螖H鈦(CaO(s)) = -635.1 kJ/mol 螖H鈦(HCl(g)) = -92.3 kJ/mol 螖H鈦(CaCl2(s)) = -795.4 kJ/mol 螖H鈦(H2O(g)) = -241.8 kJ/mol Next, we will apply the formula: 螖H鈦(reaction) = 危 螖H鈦(products) - 危 螖H鈦(reactants) 螖H鈦(reaction) = [螖H鈦(CaCl2(s)) + 螖H鈦(H2O(g))] - [螖H鈦(CaO(s)) + 2 脳 螖H鈦(HCl(g))] 螖H鈦(reaction) = [(-795.4) + (-241.8)] - [(-635.1) + 2 脳 (-92.3)] 螖H鈦(reaction) = -1037.2 + 635.1 + 184.6 螖H鈦(reaction) = -217.5 kJ/mol
02

(b) 4 FeO(s) + O2(g) -> 2 Fe2O3(s)

First, we need to find the 螖H鈦 values of the reactants and products from Appendix C. 螖H鈦(FeO(s)) = -272.0 kJ/mol 螖H鈦(O2(g)) = 0 kJ/mol (as it is an element in its standard state) 螖H鈦(Fe2O3(s)) = -824.2 kJ/mol Next, we will apply the formula: 螖H鈦(reaction) = 危 螖H鈦(products) - 危 螖H鈦(reactants) 螖H鈦(reaction) = [2 脳 螖H鈦(Fe2O3(s))] - [4 脳 螖H鈦(FeO(s)) + 螖H鈦(O2(g))] 螖H鈦(reaction) = [2 脳 (-824.2)] - [4 脳 (-272.0) + 0] 螖H鈦(reaction) = -1648.4 + 1088.0 螖H鈦(reaction) = -560.4 kJ/mol
03

(c) 2 CuO(s) + NO(g) -> Cu2O(s) + NO2(g)

First, we need to find the 螖H鈦 values of the reactants and products from Appendix C. 螖H鈦(CuO(s)) = -156.1 kJ/mol 螖H鈦(NO(g)) = 90.3 kJ/mol 螖H鈦(Cu2O(s)) = -168.6 kJ/mol 螖H鈦(NO2(g)) = 33.2 kJ/mol Next, we will apply the formula: 螖H鈦(reaction) = 危 螖H鈦(products) - 危 螖H鈦(reactants) 螖H鈦(reaction) = [螖H鈦(Cu2O(s)) + 螖H鈦(NO2(g))] - [2 脳 螖H鈦(CuO(s)) + 螖H鈦(NO(g))] 螖H鈦(reaction) = [(-168.6) + 33.2] - [2 脳 (-156.1) + 90.3] 螖H鈦(reaction) = -135.4 + 312.2 - 90.3 螖H鈦(reaction) = 86.5 kJ/mol
04

(d) 4 NH3(g) + O2(g) -> 2 N2H4(g) + 2 H2O(l)

First, we need to find the 螖H鈦 values of the reactants and products from Appendix C. 螖H鈦(NH3(g)) = -45.9 kJ/mol 螖H鈦(O2(g)) = 0 kJ/mol (as it is an element in its standard state) 螖H鈦(N2H4(g)) = 50.6 kJ/mol 螖H鈦(H2O(l)) = -285.8 kJ/mol Next, we will apply the formula: 螖H鈦(reaction) = 危 螖H鈦(products) - 危 螖H鈦(reactants) 螖H鈦(reaction) = [2 脳 螖H鈦(N2H4(g)) + 2 脳 螖H鈦(H2O(l))] - [4 脳 螖H鈦(NH3(g)) + 螖H鈦(O2(g))] 螖H鈦(reaction) = [2 脳 50.6 + 2 脳 (-285.8)] - [4 脳 (-45.9) + 0] 螖H鈦(reaction) = 101.2 - 571.6 + 183.6 螖H鈦(reaction) = -286.8 kJ/mol

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Key Concepts

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

Chemical Reactions
Chemical reactions are processes where one or more substances, called reactants, are transformed into different substances, known as products. This transformation always involves the breaking and forming of bonds, which results in changes in the energy content of the system. For example, in a reaction like
  • CaO(s) + 2HCl(g) -> CaCl鈧(s) + H鈧侽(g)
The solid calcium oxide reacts with gaseous hydrogen chloride to form solid calcium chloride and gaseous water. In chemical reactions, the laws of conservation of mass and energy are always satisfied. Mass is neither created nor destroyed, and the total energy of the system remains constant. Chemical reactions can be exothermic, releasing energy, or endothermic, absorbing energy. The classification depends on the energy required to break existing bonds compared to the energy released when new bonds are formed. Understanding these concepts helps us grasp why figuring out energy changes, like enthalpy change, is so crucial.
Thermochemistry
Thermochemistry is the branch of chemistry concerned with the heat changes that occur during chemical reactions. It allows us to predict whether a reaction will release or absorb energy. This is crucial for industrial processes, environmental science, and even daily life activities like cooking. In thermochemistry, the primary focus is on:
  • Heat (q): the energy transferred due to temperature differences.
  • Work (w): energy transfer that is not due to temperature differences, often measured as pressure-volume work.
The study of these principles allows us to understand and calculate the enthalpy ( \(\Delta H\) ), which is the heat absorbed or released under constant pressure.If a reaction is described as endothermic, it means that the system absorbs heat from its surroundings, usually resulting in a positive \(\Delta H\) . Conversely, an exothermic reaction releases heat, giving a negative \(\Delta H\) . By using thermochemical data, such as those found in Appendix C of a chemistry textbook, we can determine the heat change for reactions, like those provided in our original exercise.
Enthalpy Calculations
Enthalpy calculations are essential for predicting the energy changes in chemical reactions. Enthalpy ( \(H\) ) is a measure of the total energy of a thermodynamic system, including both internal energy and energy associated with volume and pressure. In enthalpy calculations, we use the standard enthalpy of formation ( \(\Delta H^\circ_f\) ), which is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. The formula commonly applied is: \(\Delta H^{\circ} = \Sigma \Delta H^{\circ} (\text{products}) - \Sigma \Delta H^{\circ} (\text{reactants})\). By following these steps, you can effectively determine the enthalpy change ( \(\Delta H\)) for any given reaction:
  • Find \(\Delta H^\circ_f\) values for all reactants and products from a trusted source, like Appendix C.
  • Multiply these values by their respective coefficients in the balanced chemical equation.
  • Apply the formula to find \(\Delta H^{\circ}\) by subtracting the sum of the reactants' enthalpies from the sum of the products' enthalpies.
Doing these calculations provides insight into whether a reaction is endothermic or exothermic, which in turn helps in understanding its energy profile and practical implications.

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Most popular questions from this chapter

(a) Write an equation that expresses the first law of thermodynamics in terms of heat and work. (b) Under what conditions will the quantities \(q\) and \(w\) be negative numbers?

When a \(6.50-\mathrm{g}\) sample of solid sodium hydroxide dissolves in \(100.0 \mathrm{~g}\) of water in a coffee-cup calorimeter (Figure 5.18 ), the temperature rises from \(21.6^{\circ} \mathrm{C}\) to \(37.8^{\circ} \mathrm{C}\). Calculate \(\Delta H\) (in \(\mathrm{kJ} / \mathrm{mol} \mathrm{NaOH})\) for the solution process $$ \mathrm{NaOH}(s) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ Assume that the specific heat of the solution is the same as that of pure water.

The standard enthalpies of formation of gaseous propyne \(\left(\mathrm{C}_{3} \mathrm{H}_{4}\right),\) propylene \(\left(\mathrm{C}_{3} \mathrm{H}_{6}\right),\) and propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) are +185.4 \(+20.4,\) and \(-103.8 \mathrm{~kJ} / \mathrm{mol}\), respectively. (a) Calculate the heat evolved per mole on combustion of each substance to yield \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) (b) Calculate the heat evolved on combustion of \(1 \mathrm{~kg}\) of each substance. (c) Which is the most efficient fuel in terms of heat evolved per unit mass?

Write balanced equations that describe the formation of the following compounds from elements in their standard states, and use Appendix \(\mathrm{C}\) to obtain the values of their standard enthalpies of formation: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(g)\) (b) \(\mathrm{CaCO}_{3}(s)\) (c) \(\mathrm{POCl}_{3}(l),(\mathbf{d}) \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\)

(a) Why is the change in enthalpy usually easier to measure than the change in internal energy? (b) \(H\) is a state function, but \(q\) is not a state function. Explain. (c) For a given process at constant pressure, \(\Delta H\) is positive. Is the process endothermic or exothermic?

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