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Chapter 6: Problem 50

Consider the following reaction: $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-572 \mathrm{kJ} $$ a. How much heat is evolved for the production of 1.00 mole of $\mathrm{H}_{2} \mathrm{O}(l) ?$ b. How much heat is evolved when 4.03 g hydrogen are reacted with excess oxygen? c. How much heat is evolved when 186 \(\mathrm{g}\) oxygen are reacted with excess hydrogen? d. The total volume of hydrogen gas needed to fill the Hindenburg was $2.0 \times 10^{8} \mathrm{L}\( at 1.0 atm and \)25^{\circ} \mathrm{C} .$ How much heat was evolved when the Hindenburg exploded, assuming all of the hydrogen reacted?

Short Answer

Expert verified
a. \(286\ \text{kJ}\) of heat is evolved for the production of 1.00 mole of \(H_2O (l)\). b. \(570.49\ \text{kJ}\) of heat is evolved when 4.03 g of hydrogen are reacted with excess oxygen. c. \(3323.75\ \text{kJ}\) of heat is evolved when 186 g of oxygen are reacted with excess hydrogen. d. \(2.339 \times 10^9\ \text{kJ}\) of heat was evolved when the Hindenburg exploded, assuming all of the hydrogen reacted.

Step by step solution

01

Find moles of reaction needed for 1 mole of H2O

From the balanced equation, we can see that 2 moles of H2 gas and 1 mole of O2 gas produce 2 moles of H2O. Therefore, to produce 1 mole of H2O, we need half of the reaction to take place.
02

Calculate heat evolved for half the reaction

We know that ΔH = -572 kJ for the complete reaction. As we found out in step 1, only half of the reaction is needed to produce 1 mole of H2O. Therefore, the heat evolved for the production of 1 mole of H2O would be half of -572 kJ: \[\frac{-572 \ \text{kJ}}{2} = -286\ \text{kJ}\] So, \(286\ \text{kJ}\) of heat is evolved for the production of 1.00 mole of \(H_2O (l)\). #b. Heat evolved when 4.03 g of hydrogen are reacted with excess oxygen#
03

Convert grams of H2 to moles

To find out the heat evolved, we first need to convert grams of hydrogen to moles using its molar mass: \[\text{moles of}\ H_2 = \frac{4.03 \ \text{g}}{2.02 \ \text{g/mol}} = 1.995 \ \text{moles}\]
04

Find moles of reaction

As the balanced equation shows, 2 moles of H2 react with 1 mole of O2. So, for 1.995 moles of H2, the moles of reaction would be: \[\frac{1.995\ \text{moles}}{2} = 0.9975\ \text{moles}\]
05

Calculate heat evolved for 0.9975 moles of reaction

Since ΔH = -572 kJ for the complete reaction, the heat evolved for 0.9975 moles of reaction is: \[-572 \ \text{kJ} \times 0.9975 = -570.49\ \text{kJ}\] So, \(570.49\ \text{kJ}\) of heat is evolved when 4.03 g of hydrogen are reacted with excess oxygen. #c. Heat evolved when 186 g of oxygen are reacted with excess hydrogen#
06

Convert grams of O2 to moles

To find out the heat evolved, we first need to convert grams of oxygen to moles using its molar mass: \[\text{moles of}\ O_2 = \frac{186 \ \text{g}}{32 \ \text{g/mol}} = 5.8125\ \text{moles}\]
07

Find moles of reaction

As the balanced equation shows, 1 mole of O2 reacts with 2 moles of H2. So, for 5.8125 moles of O2, the moles of reaction would be the same.
08

Calculate heat evolved for 5.8125 moles of reaction

Since ΔH = -572 kJ for the complete reaction, the heat evolved for 5.8125 moles of reaction is: \[-572 \ \text{kJ} \times 5.8125 = -3323.75\ \text{kJ}\] So, \(3323.75\ \text{kJ}\) of heat is evolved when 186 g of oxygen are reacted with excess hydrogen. #d. Heat evolved when the Hindenburg exploded#
09

Apply ideal gas equation to find moles of H2

Given volume = 2.0 × 10^8 L, pressure = 1 atm, and temperature = 25°C (298 K), we can use the ideal gas equation (PV = nRT) to find moles of H2: \[n = \frac{PV}{RT} = \frac{(2.0 \times 10^8\ \text{L})(1\ \text{atm})}{(0.0821\ \text{L atm/mol K})(298\ \text{K})} = 8.17 \times 10^6 \ \text{moles}\]
10

Find moles of reaction

As the balanced equation shows, 2 moles of H2 react with 1 mole of O2. So, for 8.17 × 10^6 moles of H2, the moles of reaction would be: \[\frac{8.17 \times 10^6\ \text{moles}}{2} = 4.085 \times 10^6\ \text{moles}\]
11

Calculate heat evolved for 4.085 × 10^6 moles of reaction

Since ΔH = -572 kJ for the complete reaction, the heat evolved for 4.085 × 10^6 moles of reaction is: \[-572\ \text{kJ} \times 4.085 \times 10^6 = -2.339 \times 10^9\ \text{kJ}\] So, \(2.339 \times 10^9\ \text{kJ}\) of heat was evolved when the Hindenburg exploded, assuming all of the hydrogen reacted.

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Most popular questions from this chapter

For each of the following situations a-c, use choices i-iii to complete the statement: "The final temperature of the water should be.." i. between \(50^{\circ} \mathrm{C}\) and \(90^{\circ} \mathrm{C}\) . ii. \(50^{\circ} \mathrm{C}\) . iii. between \(10^{\circ} \mathrm{C}\) and \(50^{\circ} \mathrm{C}\) . a. 100.0 -g sample of water at \(90^{\circ} \mathrm{C}\) is added to a 100.0 -g sample of water at \(10^{\circ} \mathrm{C}\) . b. A 100.0 -g sample of water at \(90^{\circ} \mathrm{C}\) is added to a $500.0 . \mathrm{g}\( sample of water at \)10^{\circ} \mathrm{C} .$ c. You have a Styrofoam cup with 50.0 \(\mathrm{g}\) of water at $10^{\circ} \mathrm{C}\( . You add a 50.0 -g iron ball at \)90^{\circ} \mathrm{C}$ to the water.

Which of the following substances have an enthalpy of formation equal to zero? a. \(C l_{2}(g)\) b. \(\mathrm{H}_{2}(g)\) c. \(\mathrm{N}_{2}(l)\) d. \(\mathrm{Cl}(g)\)

A gaseous hydrocarbon reacts completely with oxygen gas to form carbon dioxide and water vapor. Given the following data, determine \(\Delta H_{f}^{\circ}\) for the hydrocarbon: $$ \begin{aligned} \Delta H_{\mathrm{reacion}}^{\circ} &=-2044.5 \mathrm{kJ} / \mathrm{mol} \text { hydrocarbon } \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{CO}_{2}\right) &=-393.5 \mathrm{kJ} / \mathrm{mol} \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right) &=-242 \mathrm{kJ} / \mathrm{mol} \end{aligned} $$ Density of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) product mixture at 1.00 \(\mathrm{atm}\) , \(200 . \mathrm{C}=0.751 \mathrm{g} / \mathrm{L}\) . The density of the hydrocarbon is less than the density of Kr at the same conditions.

The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area \((1 \text { watt }=1 \mathrm{Js} \text { ). The plants in an }\) agricultural field produce the equivalent of \(20 . \mathrm{kg}\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per hectare \(\left(1 \mathrm{ha}=10,000 \mathrm{m}^{2}\right) .\) Assuming that sucrose is produced by the reaction $$ \begin{aligned} 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+& 12 \mathrm{O}_{2}(g) \\ & \Delta H=5640 \mathrm{kJ} \end{aligned} $$ calculate the percentage of sunlight used to produce the sucrose-that is, determine the efficiency of photosynthesis.

In which of the following systems is (are) work done by the surroundings on the system? Assume pressure and temperature are constant. a. $2 \operatorname{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \operatorname{SO}_{3}(g)$ b. \(\mathrm{CO}_{2}(s) \longrightarrow \mathrm{CO}_{2}(g)\) c. $4 \mathrm{NH}_{3}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$ d. \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) e. \(\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaCO}(s)+\mathrm{CO}_{2}(g)\)
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