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Chapter 1: Problem 20

The enthalpy change for which of the following reactions would be equal to the enthalpy of formation for ethanol $\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right)$ ? (A) $\mathrm{CH}_{3}+\mathrm{CH}_{2}+\mathrm{OH} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ (B) $2 \mathrm{C}+5 \mathrm{H}+\mathrm{O} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ (C) $4 \mathrm{C}+6 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ (D) $2 \mathrm{C}+3 \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$

Short Answer

Expert verified
The enthalpy change for the reaction as defined in option (D) would be equal to the enthalpy of formation for ethanol.

Step by step solution

01

Understanding Enthalpy of Formation

The enthalpy of formation of a compound is the change in enthalpy that accompanies the formation of 1 mole of the compound from its elements, with all substances in their standard states at 1 atm. The substances in their standard states refers to the most stable form of the substances under normal conditions (1 atm pressure and room temperature).
02

Analyze Given Options

For each of the options, analyze the reactants to see if they are in their standard states: \[ (A) CH_{3} + CH_{2} + OH \] These are not the standard states of Carbon, Hydrogen, and Oxygen. \[ (B) 2C + 5H + O \] Hydrogen and Oxygen are not in their standard states here. Hydrogen in its standard state is H2 (hydrogen molecule) and Oxygen in its standard state is O2 (oxygen molecule). \[ (C) 4C + 6H2 + O2 \] Although the elements are in their standard states, the reaction gives 2 moles of ethanol. The enthalpy of formation should correspond to the formation of 1 mole of the compound. \[ (D) 2C + 3H2 + 0.5O2 \] This reaction involves the elements in their standard states and forms one mole of ethanol. Therefore, this is the correct answer.
03

Selection of Answer

After analyzing all the given options, it is clear that option (D) provides the correct stoichiometry for the enthalpy of formation for ethanol. The reaction contains all the elements required for the formation of ethanol in their standard states and forms one mole of ethanol, which is consistent with the definition of enthalpy of formation.

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Most popular questions from this chapter

150 \(\mathrm{mL}\) of saturated \(\mathrm{SrF}_{2}\) solution is present in a 250 \(\mathrm{mL}\) beaker at room temperature. The molar solubility of \(\mathrm{SrF}_{2}\) at 298 \(\mathrm{K}\) is \(1.0 \times 10^{-3} \mathrm{M}\) . What are the concentrations of \(\mathrm{Sr}^{2+}\) and \(\mathrm{F}^{-}\) in the beaker? (A) \(\left[\mathrm{Sr}^{2+}\right]=1.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-3} M\) (B) \(\left[\mathrm{Sr}^{2+}\right]=1.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=2.0 \times 10^{-3} M\) (C) \(\left[\mathrm{Sr}^{2+}\right]=2.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-3} M\) (D) \(\left[\mathrm{Sr}^{2+}\right]=2.0 \times 10^{-3} \mathrm{M}\left[\mathrm{F}^{-}\right]=2.0 \times 10^{-3} M\)

A strong acid/strong base titration is completed using an indicator which changes color at the exact equivalence point of the titration. The protonated form of the indicator is HIn, and the deprotonated form is In. At the equivalence point of the reaction: (A) \([\mathrm{HIn}]=\left[\mathrm{In}^{-}\right]\) (B) \([\mathrm{HIn}]=1 /\left[\mathrm{In}^{-}\right]\) (C) \([\mathrm{HIn}]=2\left[\mathrm{In}^{-}\right]\) (D) \([\mathrm{HIn}]=\left[\mathrm{In}^{-}\right]^{2}\)

\(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightarrow 2 \mathrm{HF}(g)\) 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 \(\mathrm{kJ} / \mathrm{mol}\)

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ If the \(\mathrm{Cu}^{+}+e^{-} \rightarrow \mathrm{Cu}(s)\) half-reaction has a standard reduction potential of \(+0.52 \mathrm{V},\) what is the standard reduction potential for the \(\mathrm{Zn}^{2+}+2 e^{-} \rightarrow \mathrm{Zn}(s)\) half-reaction? (A) \(+0.76 \mathrm{V}\) (B) \(-0.76 \mathrm{V}\) (C) \(+0.24 \mathrm{V}\) (D) \(-0.24 \mathrm{V}\)

The following reaction is found to be at equilibrium at 25°C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) What is the expression for the equilibrium constant, \(K_{\mathrm{c}} ?\) (A) \(\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]\left[\mathrm{SO}_{2}\right]^{2}}\) (B) \(\frac{2\left[\mathrm{SO}_{3}\right]}{\left[\mathrm{O}_{2}\right] 2\left[\mathrm{SO}_{2}\right]}\) (C) \(\frac{\left[\mathrm{O}_{2}\right]\left[\mathrm{SO}_{2}\right]^{2}}{\left[\mathrm{SO}_{3}\right]^{2}}\) (D) \(\frac{\left[\mathrm{O}_{2}\right] 2\left[\mathrm{SO}_{2}\right]}{2\left[\mathrm{SO}_{3}\right]}\)
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