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Chapter 5: Problem 25

Calculate \(\Delta E\), and determine whether the process is endothermic or exothermic for the following cases: (a) A system absorbs \(105 \mathrm{~kJ}\) of heat from its surroundings while doing \(29 \mathrm{~kJ}\) of work on the surroundings; (b) \(q=1.50 \mathrm{~kJ}\) and \(w=-657 \mathrm{~J} ;\) (c) the system releases \(57.5 \mathrm{~kJ}\) of heat while doing \(22.5 \mathrm{~kJ}\) of work on the surroundings.

Short Answer

Expert verified
(a) \(\Delta E = 134 \mathrm{~kJ}\) (endothermic); (b) \(\Delta E = 0.843 \mathrm{~kJ}\) (endothermic); (c) \(\Delta E = -35 \mathrm{~kJ}\) (exothermic)

Step by step solution

01

Identify the Given Values

For this case, q = +105 kJ (absorbed) and w = +29 kJ (work done on surroundings).
02

Calculate 螖E for Case (a)

Use the formula, 螖E = q + w: 螖E = (+105 kJ) + (+29 kJ) = 134 kJ
03

Determine if the Process is Endothermic or Exothermic

Since 螖E is positive, the process is endothermic. Case (b):
04

Identify the Given Values

For this case, q = +1.50 kJ (absorbed) and w = -657 J (work done by the system). Convert w to kJ: w = -657 J * (1 kJ / 1000 J) = -0.657 kJ
05

Calculate 螖E for Case (b)

Use the formula, 螖E = q + w: 螖E = (+1.50 kJ) + (-0.657 kJ) = 0.843 kJ
06

Determine if the Process is Endothermic or Exothermic

Since 螖E is positive, the process is endothermic. Case (c):
07

Identify the Given Values

For this case, q = -57.5 kJ (released) and w = +22.5 kJ (work done on surroundings).
08

Calculate 螖E for Case (c)

Use the formula, 螖E = q + w: 螖E = (-57.5 kJ) + (+22.5 kJ) = -35 kJ
09

Determine if the Process is Endothermic or Exothermic

Since 螖E is negative, the process is exothermic.

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Most popular questions from this chapter

Consider the following reaction, which occurs at room temperature and pressure: $$ 2 \mathrm{Cl}(g) \longrightarrow \mathrm{Cl}_{2}(g) \quad \Delta H=-243.4 \mathrm{~kJ} $$ Which has the higher enthalpy under these conditions, \(2 \mathrm{Cl}(g)\) or \(\mathrm{Cl}_{2}(g) ?\)

Consider the combustion of liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}(I)\) : \(\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\Delta H=-726.5 \mathrm{~kJ}\) (a) What is the enthalpy change for the reverse reaction? (b) Balance the forward reaction with whole-number coefficients. What is \(\Delta H\) for the reaction represented by this equation? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If the reaction were written to produce \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) instead of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\), would you expect the magni- tude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

Many cigarette lighters contain liquid butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\). Using standard enthalpies of formation, calculate the quantity of heat produced when \(5.00 \mathrm{~g}\) of butane is completely combusted in air under standard conditions.

From the enthalpies of reaction calculate \(\Delta H\) for the reaction of ethylene with \(\mathrm{F}_{2}\) $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+6 \mathrm{~F}_{2}(g)-\cdots 2 \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g) $$

Gasoline is composed primarily of hydrocarbons, including many with eight carbon atoms, called octanes. One of the cleanest-burning octanes is a compound called \(2,3,4\) -trimethylpentane, which has the following structural formula: The complete combustion of one mole of this compound to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) leads to \(\Delta H^{\circ}=-5064.9 \mathrm{~kJ} / \mathrm{mol}\). (a) Write a balanced equation for the combustion of 1 mol of \(\mathrm{C}_{8} \mathrm{H}_{18}(l) .(\mathrm{b})\) Write a balanced equation for the formation of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) from its elements. (c) By using the information in this problem and data in Table \(5.3\), calculate \(\Delta H_{f}^{\circ}\) for \(2,3,4\) -trimethylpentane.
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