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Chapter 5: Problem 103

The corrosion (rusting) of iron in oxygen-free water includes the formation of iron(II) hydroxide from iron by the following reaction: $$ \mathrm{Fe}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2}(g) $$ If 1 mol of iron reacts at \(298 \mathrm{~K}\) under \(101.3 \mathrm{kPa}\) pressure, the reaction performs \(2.48 \mathrm{~J}\) of \(P-V\) work, pushing back the atmosphere as the gaseous \(\mathrm{H}_{2}\) forms. At the same time, $11.73 \mathrm{~kJ}$ of heat is released to the environment. What are the values of \(\Delta H\) and of \(\Delta E\) for this reaction?

Short Answer

Expert verified
The change in enthalpy (ΔH) for this reaction is approximately \(-11,480.548 \: J\) and the change in internal energy (ΔE) is \(-11,727.52 \: J\).

Step by step solution

01

Calculate the work done

Given that the reaction performs 2.48 J of P-V work, we have w = 2.48 J.
02

Calculate the heat released

Given that 11.73 kJ of heat is released to the environment, we have q = -11.73 kJ (the negative sign indicates heat is released). To keep the units consistent, convert q to J: q = -11.73 kJ * 1000 J/kJ = -11730 J.
03

Calculate the change in internal energy (ΔE)

Using the relation ΔE = q + w, ΔE = (-11730 J) + (2.48 J) = -11727.52 J
04

Calculate the change in volume (ΔV)

Given that 1 mol of iron reacts, we have 1 mol of H₂ gas produced. Using the ideal gas law, PV = nRT, with pressure P = 101.3 kPa, n = 1 mol, R = 8.314 J/(mol*K), and temperature T = 298 K, ΔV = nRT/P = (1 mol) * (8.314 J/(mol*K)) * (298 K) / (101.3 kPa * (1 kJ/(kJ * J))) = 0.0244 m³/mol
05

Calculate ΔH

Using the relation ΔH = ΔE + PΔV, ΔH = (-11727.52 J) + (101.3 kPa * 0.0244 m³/mol) * (1 J/(1 kPa * m³)) ΔH = -11727.52 J + 246.972 J = -11480.548 J (approximately) Thus, the change in enthalpy (ΔH) for this reaction is approximately -11480.548 J and the change in internal energy (ΔE) is -11727.52 J.

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Most popular questions from this chapter

The gas-phase reaction shown, between \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\), was run in an apparatus designed to maintain a constant pressure. (a) Write a balanced chemical equation for the reaction depicted and predict whether \(w\) is positive, negative, or zero. (b) Using data from Appendix C, determine \(\Delta H\) for the formation of one mole of the product. [Sections 5.3 and 5.7\(]\)

The complete combustion of methane, \(\mathrm{CH}_{4}(g)\), to form \(\mathrm{H}_{2} \mathrm{O}(l)\) and \(\mathrm{CO}_{2}(g)\) at constant pressure releases \(890 \mathrm{~kJ}\) of heat per mole of \(\mathrm{CH}_{4}\). (a) Write a balanced thermochemical equation for this reaction. (b) Draw an enthalpy diagram for the reaction.

Without doing any calculations, predict the sign of \(\Delta H\) for each of the following reactions: (a) $\mathrm{NaCl}(s) \longrightarrow \mathrm{Na}^{+}(g)+\mathrm{Cl}^{-}(\mathrm{g})$ (b) \(2 \mathrm{H}(g) \longrightarrow \mathrm{H}_{2}(g)\) (c) \(\mathrm{Na}(g) \longrightarrow \mathrm{Na}^{+}(g)+\mathrm{e}^{-}\) (d) \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(l)\)

The hydrocarbons cyclohexane $\left(\mathrm{C}_{6} \mathrm{H}_{12}(l), \Delta H_{f}^{\circ}=-156\right.$ \(\mathrm{kJ} / \mathrm{mol}\) ) and 1-hexene $\left(\mathrm{C}_{6} \mathrm{H}_{12}(l), \Delta H_{f}^{\circ}=-74 \mathrm{~kJ} / \mathrm{mol}\right)$ have the same empirical formula. (a) Calculate the standard enthalpy change for the transformation of cyclohexane to 1-hexene. (b) Which has greater enthalpy, cyclohexane or 1-hexene? (c) Without doing a further calculation and knowing the answer to (b), do you expect cyclohexane or 1-hexene to have the larger combustion enthalpy?

(a) What amount of heat (in joules) is required to raise the temperature of $1 \mathrm{~g}$ of water by 1 kelvin? (b) What amount of heat (in joules) is required to raise the temperature of 1 mole of water by 1 kelvin? (c) What is the heat capacity of \(370 \mathrm{~g}\) of liquid water? (d) How many kJ of heat are needed to raise the temperature of $5.00 \mathrm{~kg}\( of liquid water from 24.6 to \)46.2^{\circ} \mathrm{C} ?$
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