Chapter 10: Problem 15
What do we mean by thermodynamics? What is the first law of thermodynamics?
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 10: Problem 15
What do we mean by thermodynamics? What is the first law of thermodynamics?
These are the key concepts you need to understand to accurately answer the question.
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Get started for freeCalculate \(\Delta H\) for the reaction $$\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ given the following data: \(\begin{array}{ll}\text { Equation } & \Delta H(\mathrm{~kJ}) \\ 2 \mathrm{NH}_{3}(g)+3 \mathrm{~N}_{2} \mathrm{O}(g) \rightarrow 4 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) & -1010 \\\ \mathrm{~N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & -317 \\ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & -143 \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)\end{array}\)
The combustion of methane, \(\mathrm{CH}_{4}\), is an exothermic process. Therefore, the products of this reaction must possess (higher/lower) total potential energy than do the reactants.
Students often confuse what is meant by heat and temperature. Define each. How are the two concepts related?
Consider the reaction $$\mathrm{B}_{2} \mathrm{H}_{6}(g)+3 \mathrm{O}_{2}(g) \rightarrow \mathrm{B}_{2} \mathrm{O}_{3}(s)+3\mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=-2035 \mathrm{~kJ}$$ Calculate the amount of heat released when \(54.0 \mathrm{~g}\) of diborane is combusted.
Consider the following equations: \(\begin{aligned} 3 \mathrm{~A}+6 \mathrm{~B} \rightarrow 3 \mathrm{D} & & \Delta H=-403 \mathrm{~kJ} / \mathrm{mol} \\\ \mathrm{E}+2 \mathrm{~F} \rightarrow \mathrm{A} & & \Delta H=-105.2 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{C} \rightarrow \mathrm{E}+3 \mathrm{D} & & \Delta H=+64.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned}\) Suppose the first equation is reversed and multiplied by \(\frac{1}{6},\) the second and third equations are divided by \(2,\) and the three adjusted equations are added. What is the net reaction and what is the overall heat of this reaction?
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