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Chapter 6: Problem 16

For the reaction \(\mathrm{HgO}(s) \rightarrow \mathrm{Hg}(l)+\frac{1}{2} \mathrm{O}_{2}(g), \Delta H=+90.7 \mathrm{~kJ}:\) a. What quantity of heat is required to produce \(1 \mathrm{~mol}\) of mercury by this reaction? b. What quantity of heat is required to produce \(1 \mathrm{~mol}\) of oxygen gas by this reaction? c. What quantity of heat would be released in the following reaction as written? $$ 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{HgO}(s) $$

Short Answer

Expert verified
Answer: a) +90.7 kJ is required to produce 1 mol of Hg. b) +181.4 kJ is required to produce 1 mol of O₂. c) -181.4 kJ of heat would be released in the reverse reaction as written.

Step by step solution

01

Given information

We have the balanced equation: HgO(s) → Hg(l) + 0.5 O₂(g) with ΔH = +90.7 kJ.
02

Calculate heat for 1 mol Hg

In this reaction, 1 mol of HgO decomposes to produce 1 mol of Hg. Thus, the heat required to produce 1 mol Hg is directly given by the value of ΔH, which is +90.7 kJ. Answer (a): +90.7 kJ is required to produce 1 mol of Hg. #b. Calculating heat for 1 mol O₂ production#
03

Determine stoichiometric relationships

From the balanced equation: HgO(s) → Hg(l) + 0.5 O₂(g), we can see that 2 mol of HgO are needed to produce 1 mol of O₂.
04

Calculate heat for 1 mol O₂

Since ΔH is +90.7 kJ for 1 mol HgO producing 0.5 mol O₂, in order to produce 1 mol O₂, we need to double this value. Therefore, the heat required to produce 1 mol O₂ is: \(2 \times 90.7 \text{ kJ} = 181.4 \text{ kJ}\) Answer (b): +181.4 kJ is required to produce 1 mol of O₂. #c. Calculating heat released in the reverse reaction#
05

Reverse reaction

The reverse reaction is: $$ 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{HgO}(s) $$
06

Calculate heat released in the reverse reaction

Since the reaction is reversed and we are dealing with the inverse stoichiometry, the heat change will also have an opposite sign. (Note that the forward reaction ΔH = +90.7 kJ for 1 mol HgO) Thus, for 2 mol HgO produced in the reverse reaction, the heat change would be: $$ \Delta H_{\text{reverse}} = -1 \times (2 \times 90.7 \text{ kJ}) = -181.4 \text{ kJ} $$ Answer (c): -181.4 kJ of heat would be released in the reverse reaction as written.

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Most popular questions from this chapter

Write reactions for which the enthalpy change will be a. \(\Delta H_{\mathrm{f}}^{\circ}\) for solid aluminum oxide. b. The standard enthalpy of combustion of liquid ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) .\) c. The standard enthalpy of neutralization of sodium hydroxide solution by hydrochloric acid. d. \(\Delta H_{\mathrm{f}}^{\circ}\) for gaseous vinyl chloride, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}(g)\). e. The enthalpy of combustion of liquid benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\). f. The enthalpy of solution of solid ammonium bromide.

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 \(40.6 \mathrm{~kJ} / \mathrm{mol}\).)

On Easter Sunday, April 3, 1983 , nitric acid spilled from a tank car near downtown Denver, Colorado. The spill was neutralized with sodium carbonate: \(2 \mathrm{HNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(s) \longrightarrow 2 \mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) a. Calculate \(\Delta H^{\circ}\) for this reaction. Approximately \(2.0 \times 10^{4}\) gal nitric acid was spilled. Assume that the acid was an aqueous solution containing \(70.0 \% \mathrm{HNO}_{3}\) by mass with a density of \(1.42 \mathrm{~g} / \mathrm{cm}^{3}\). What mass of sodium carbonate was required for complete neutralization of the spill, and what quantity of heat was evolved? \(\left(\Delta H_{\mathrm{f}}^{\circ}\right.\) for \(\mathrm{NaNO}_{3}(a q)=-467 \mathrm{~kJ} / \mathrm{mol}\) ) b. According to The Denver Post for April 4, 1983 , authorities feared that dangerous air pollution might occur during the neutralization. Considering the magnitude of \(\Delta H^{\circ}\), what was their major concern?

When \(1.00 \mathrm{~L}\) of \(2.00 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution at \(30.0^{\circ} \mathrm{C}\) is added to \(2.00 \mathrm{~L}\) of \(0.750 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) solution at \(30.0^{\circ} \mathrm{C}\) in a calorimeter, a white solid \(\left(\mathrm{BaSO}_{4}\right)\) forms. The temperature of the mixture increases to \(42.0^{\circ} \mathrm{C}\). Assuming that the specific heat capacity of the solution is \(6.37 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and that the density of the final solution is \(2.00 \mathrm{~g} / \mathrm{mL}\), calculate the enthalpy change per mole of \(\mathrm{BaSO}_{4}\) formed.

The standard enthalpy of combustion of ethene gas, \(\mathrm{C}_{2} \mathrm{H}_{4}(g)\), is \(-1411.1 \mathrm{~kJ} / \mathrm{mol}\) at \(298 \mathrm{~K}\). Given the following enthalpies of formation, calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{C}_{2} \mathrm{H}_{4}(g)\). $$ \begin{array}{ll} \mathrm{CO}_{2}(g) & -393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2} \mathrm{O}(l) & -285.8 \mathrm{~kJ} / \mathrm{mol} \end{array} $$
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