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Chapter 5: Problem 71

Using values from Appendix \(C\), calculate the standard enthalpy change for each of the following reactions: (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{Mg}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{MgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g)+4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{SiCl}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{SiO}_{2}(s)+4 \mathrm{HCl}(g)\)

Short Answer

Expert verified
The standard enthalpy changes for the given reactions are: (a) 螖H = -197.8 \(kJ/mol\) (b) 螖H = 36.9 \(kJ/mol\) (c) 螖H = -976.5 \(kJ/mol\) (d) 螖H = -250.5 \(kJ/mol\)

Step by step solution

01

Find the Enthalpy of Formation Values

Obtain the standard enthalpies of formation (螖H_f) from Appendix C or a similar source for SO_2(g), O_2(g), and SO_3(g): - 螖H_f(SO_2,g) = -296.8 \(kJ/mol\) - 螖H_f(O_2,g) = 0 \(kJ/mol\) (this is because O2 is in its standard state) - 螖H_f(SO_3,g) = -395.7 \(kJ/mol\)
02

Calculate the Enthalpy Change

Apply the formula for 螖H_reaction: 螖H_reaction = [(2)*(-395.7) - (2)*(-296.8) - (1)*(0)] \(kJ/mol\) = -197.8 \(kJ/mol\) #(b) Mg(OH)_2(s) -> MgO(s) + H_2O(l)#
03

Find the Enthalpy of Formation Values

Obtain the standard enthalpies of formation for Mg(OH)_2(s), MgO(s), and H_2O(l): - 螖H_f(Mg(OH)_2,s) = -924.7 \(kJ/mol\) - 螖H_f(MgO,s) = -601.6 \(kJ/mol\) - 螖H_f(H_2O,l) = -285.8 \(kJ/mol\)
04

Calculate the Enthalpy Change

Apply the formula for 螖H_reaction: 螖H_reaction = [(1)*(-601.6) + (1)*(-285.8) - (1)*(-924.7)] \(kJ/mol\) = 36.9 \(kJ/mol\) #(c) N_2O_4(g) + 4 H_2(g) -> N_2(g) + 4 H_2O(g)#
05

Find the Enthalpy of Formation Values

Obtain the standard enthalpies of formation for N_2O_4(g), H_2(g), N_2(g), and H_2O(g): - 螖H_f(N_2O_4,g) = 9.7 \(kJ/mol\) - 螖H_f(H_2,g) = 0 \(kJ/mol\) (this is because H2 is in its standard state) - 螖H_f(N_2,g) = 0 \(kJ/mol\) (this is because N2 is in its standard state) - 螖H_f(H_2O,g) = -241.8 \(kJ/mol\)
06

Calculate the Enthalpy Change

Apply the formula for 螖H_reaction: 螖H_reaction = [(1)*(0) + 4*(-241.8) - (1)*(9.7) - 4*(0)] \(kJ/mol\) = -976.5 \(kJ/mol\) #(d) SiCl_4(l) + 2 H_2O(l) -> SiO_2(s) + 4 HCl(g)#
07

Find the Enthalpy of Formation Values

Obtain the standard enthalpies of formation for SiCl_4(l), H_2O(l), SiO_2(s), and HCl(g): - 螖H_f(SiCl_4,l) = -640.1 \(kJ/mol\) - 螖H_f(H_2O,l) = -285.8 \(kJ/mol\) - 螖H_f(SiO_2,s) = -910.9 \(kJ/mol\) - 螖H_f(HCl,g) = -92.3 \(kJ/mol\)
08

Calculate the Enthalpy Change

Apply the formula for 螖H_reaction: 螖H_reaction = [(1)*(-910.9) + 4*(-92.3) - (1)*(-640.1) - 2*(-285.8)] \(kJ/mol\) = -250.5 \(kJ/mol\) The standard enthalpy changes for the given reactions are: -(a) 螖H = -197.8 \(kJ/mol\) -(b) 螖H = 36.9 \(kJ/mol\) -(c) 螖H = -976.5 \(kJ/mol\) -(d) 螖H = -250.5 \(kJ/mol\)

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Most popular questions from this chapter

Does \(\Delta H_{\mathrm{rxn}}\) for the reaction represented by the following equation equal the standard enthalpy of formation for \(\mathrm{CH}_{3} \mathrm{OH}(l) ?\) Why or why not? [Section 5.7] $$ \mathrm{C}(\text { graphite })+4 \mathrm{H}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l) $$

From the enthalpies of reaction \(\begin{aligned} 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) & \Delta H &=-483.6 \mathrm{~kJ} \\ 3 \mathrm{O}_{2}(g) & \cdots &-\cdots & 2 \mathrm{O}_{3}(g) & \Delta H &=+284.6 \mathrm{~kJ} \end{aligned}\) calculate the heat of the reaction $$ 3 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{3}(\mathrm{~g}) \longrightarrow 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$

(a) Calculate the kinetic energy in joules of a \(45-\mathrm{g}\) golf ball moving at \(61 \mathrm{~m} / \mathrm{s}\). (b) Convert this energy to calories. (c) What happens to this energy when the ball lands in a sand trap?

A coffee-cup calorimeter of the type shown in Figure \(5.17\) contains \(150.0 \mathrm{~g}\) of water at \(25.1^{\circ} \mathrm{C}\). A \(121.0\) -g block of copper metal is heated to \(100.4^{\circ} \mathrm{C}\) by putting it in a beaker of boiling water. The specific heat of \(\mathrm{Cu}(s)\) is \(0.385 \mathrm{~J} / \mathrm{g}-\mathrm{K} .\) The \(\mathrm{Cu}\) is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of \(30.1^{\circ} \mathrm{C}\). (a) Determine the amount of heat, in \(J\), lost by the copper block. (b) Determine the amount of heat gained by the water. The specific heat of water is \(4.18 \mathrm{~J} / \mathrm{g}-\mathrm{K} .\) (c) The difference between your answers for (a) and (b) is due to heat loss through the Styrofoam \(^{8}\) cups and the heat necessary to raise the temperature of the inner wall of the apparatus. The heat capacity of the calorimeter is the amount of heat necessary to raise the temperature of the apparatus (the cups and the stopper) by \(1 \mathrm{~K}\). Calculate the heat capacity of the calorimeter in J/K. (d) What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter?

A sample of a hydrocarbon is combusted completely in \(\mathrm{O}_{2}(g)\) to produce \(21.83 \mathrm{~g} \mathrm{CO}_{2}(g), 4.47 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), and \(311 \mathrm{~kJ}\) of heat. (a) What is the mass of the hydrocarbon sample that was combusted? (b) What is the empirical formula of the hydrocarbon? (c) Calculate the value of \(\Delta H_{f}^{\circ}\) per empirical-formula unit of the hydrocarbon. (d) Do you think that the hydrocarbon is one of those listed in Appendix C? Explain your answer.
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