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Chapter 1: Problem 19

\(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightarrow 2 \mathrm{HF}(g)\) Gaseous hydrogen and fluorine combine in the reaction above to form hydrogen fluride with an enthalpy change of \(-540 \mathrm{kJ}\) . What is the value of the heat of formation of \(\mathrm{HF}(g) ?\) (A) \(-1,080 \mathrm{kJ} / \mathrm{mol}\) (B) \(-270 \mathrm{kJ} / \mathrm{mol}\) (C) 270 \(\mathrm{kJ} / \mathrm{mol}\) (D) 540 \(\mathrm{kJ} / \mathrm{mol}\)

Short Answer

Expert verified
The heat of formation for HF(g) is -270 kJ/mol, so the correct answer is (B) -270 kJ/mole.

Step by step solution

01

Determine the reaction stoichiometry

The reaction shows that 1 mole of H2(g) reacts with 1 mole of F2(g) to produce 2 moles of HF(g). Therefore, the amount of energy (-540 kJ) associated with this reaction is for the formation of 2 moles of HF(g).
02

Calculate the heat of formation for one mole of HF

The heat of formation is the energy associated with forming one mole of a substance. Since the given reaction forms 2 moles of HF(g), we divide the total energy by 2 to find the heat of formation for one mole of HF(g). Doing this gives us -540 kJ / 2 = -270 kJ/mole.

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Most popular questions from this chapter

When calcium chloride \(\left(\mathrm{CaCl}_{2}\right)\) dissolves in water, the temperature of the water increases dramatically. During this reaction, heat transfers from (A) the reactants to the products (B) the reactants to the system (C) the system to the surroundings (D) the products to the surroundings

A student titrates 20.0 \(\mathrm{mL}\) of 1.0 \(M \mathrm{NaOH}\) with 2.0 \(\mathrm{M}\), \(\mathrm{HCO}_{2} \mathrm{H}\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right) .\) Formic acid is a monoprotic acid. \(\mathrm{CH}_{3} \mathrm{NH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \leftrightarrow \mathrm{OH}^{-}(a q)+\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}(a q)\) The above equation represents the reaction between the base methylamine \(\left(K_{\mathrm{b}}=4.38 \times 10^{-4}\right)\) and water. Which of the following best represents the.concentrations of the various species at equilibrium? (A) \(\left[\mathrm{OH}^{-}\right]>\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]\) (B) \(\left[\mathrm{OH}^{-}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]\) (C) \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]>\left[\mathrm{OH}^{-}\right]>\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]\) (D) \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]>\left[\mathrm{OH}^{-}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}+\right]\)

The following mechanism is proposed for a reaction: \(\begin{array}{ll}{2 \mathrm{A} \rightarrow \mathrm{B}} & {\text { (fast equilibrium) }} \\ {\mathrm{C}+\mathrm{B} \rightarrow \mathrm{D}} & {\text { (slow) }} \\ {\mathrm{D}+\mathrm{A} \rightarrow \mathrm{E}} & {\text { (fast) }}\end{array}\) Which of the following is the correct rate law for compete reaction? (A) Rate \(=k[\mathrm{C}]^{2}[\mathrm{B}]\) (B) Rate \(=k\left[\mathrm{Cl}[\mathrm{A}]^{2}\right.\) (C) Rate \(=k[\mathrm{C}][\mathrm{A}]^{3}\) (D) Rate \(=k[\mathrm{D}][\mathrm{A}]\)

$$4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ The above reaction will experience a rate increase by the addition of a catalyst such as platinum. Which of the following best explains why? (A) The catalyst causes the value for \(\Delta G\) to become more negative. (B) The catalyst increases the percentage of collisions that occur at the proper orientation in the reactant molecules. (C) The catalyst introduces a new reaction mechanism for the reaction. (D) The catalyst increases the activation energy for the reaction.

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: $\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}$ The concentrations of the solutions in each half-cell are 1.0 M. Which net ionic equation below represents a possible reaction that takes place when a strip of magnesium metal is oxidized by a solution of chromium (III) nitrate? (A) $\operatorname{Mg}(s)+\operatorname{Cr}\left(\mathrm{NO}_{3}\right)_{3}(a q) \rightarrow \mathrm{Mg}^{2+}(a q)+\mathrm{Cr}^{3+}(a q)+3 \mathrm{NO}_{3}^{-}(a q)$ (B) $3 \mathrm{Mg}(s)+2 \mathrm{Cr}^{3+} \rightarrow 3 \mathrm{Mg}^{2+}+2 \mathrm{Cr}(s)$ (C) $\mathrm{Mg}(s)+\mathrm{Cr}^{3+} \rightarrow \mathrm{Mg}^{2+}+\mathrm{Cr}(s)$ (D) $3 \mathrm{Mg}(s)+2 \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}(a q) \rightarrow 3 \mathrm{Mg}^{2+}(a q)+2 \mathrm{Cr}(s)+\mathrm{NO}_{3}^{-}(a q)$
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