/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 When 1 mol of iron metal reacts ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When 1 mol of iron metal reacts with hydrochloric acid at constant temperature and pressure to produce hydrogen gas and aqueous iron(II) chloride, \(89.1 \mathrm{~kJ}\) of heat evolves. Write a thermochemical equation for this reaction.

Short Answer

Expert verified
Fe (s) + 2 HCl (aq) → H₂ (g) + FeCl₂ (aq); ΔH = -89.1 kJ.

Step by step solution

01

Write the Balanced Chemical Equation

Identify the reactants and products of the reaction. Iron (Fe) reacts with hydrochloric acid (HCl) to produce hydrogen gas (Hâ‚‚) and aqueous iron(II) chloride (FeClâ‚‚). The balanced chemical equation for this reaction is: \[ \text{Fe (s)} + 2 \text{HCl (aq)} \rightarrow \text{H}_2 (g) + \text{FeCl}_2 (aq) \]
02

Identify the Amount of Heat Evolved and its Sign

The problem states that 89.1 kJ of heat is evolved. This implies that the reaction is exothermic. When the reaction is exothermic, the heat has a negative sign, indicating that heat is released. Therefore, the heat associated with the reaction is \(-89.1 \mathrm{~kJ}\).
03

Write the Thermochemical Equation

Combine the balanced chemical equation with the heat change to form the thermochemical equation. Here, we include the heat change value in kJ beside the equation, transferred to the product side because it is exothermic. The thermochemical equation is: \[ \text{Fe (s)} + 2 \text{HCl (aq)} \rightarrow \text{H}_2 (g) + \text{FeCl}_2 (aq); \Delta H = -89.1 \mathrm{~kJ} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Balanced Chemical Equation
A balanced chemical equation is essential to accurately represent what happens in a chemical reaction. It ensures that the number of atoms of each element is the same on both the reactant and product sides, complying with the Law of Conservation of Mass. For the reaction where iron reacts with hydrochloric acid producing hydrogen gas and aqueous iron(II) chloride, we need to balance it correctly. This reaction involves the following components:

  • Reactants: Solid iron (Fe) and aqueous hydrochloric acid (HCl)
  • Products: Hydrogen gas (Hâ‚‚) and aqueous iron(II) chloride (FeClâ‚‚)
To balance this equation, ensure the number of atoms for each element is equal on both sides. We get the balanced equation:
\[ \text{Fe (s)} + 2 \text{HCl (aq)} \rightarrow \text{H}_2 (g) + \text{FeCl}_2 (aq) \]
By doing this, we keep the reaction formula accurate, showing the transformation from reactants to products while respecting mass conservation principles.
Exothermic Reaction
When a chemical reaction releases heat into its surroundings, it is classified as an exothermic reaction. In such reactions, the energy required to break bonds in the reactants is less than the energy released when new bonds form in the products. The reaction between iron and hydrochloric acid is exothermic, releasing a significant amount of heat.

This heat release can be observed through several indicators:
  • The temperature of the surroundings increases.
  • The system feels warm to touch.
  • The heat value is negative, indicating that energy flows out of the system, not into it.
In our specific example, the heat evolved is 89.1 kJ. Because energy is released, we assign a negative sign to this value. So, the heat change is noted as \(-89.1 \text{ kJ}\).
This characteristic helps in understanding energy flow within chemical processes and can be crucial for reactions used in energy-generating applications.
Enthalpy Change
Enthalpy change, symbolized as \(\Delta H\), is a vital thermodynamic quantity that helps quantify the heat change during a chemical reaction at constant pressure. It reflects the difference in the total heat content of products and reactants. If the enthalpy change is negative, as seen in exothermic reactions, it indicates that heat is lost from the system to the surroundings.
For our reaction involving iron and hydrochloric acid, \(\Delta H\) is \(-89.1 \text{ kJ}\), signifying an exothermic process. This value is included in the thermochemical equation alongside the balanced chemical reaction, presenting a comprehensive depiction of the reaction:
\[ \text{Fe (s)} + 2 \text{HCl (aq)} \rightarrow \text{H}_2 (g) + \text{FeCl}_2 (aq); \Delta H = -89.1 \text{ kJ} \]
Enthalpy changes help predict whether a reaction will occur spontaneously or if it necessitates additional energy inputs. They are also fundamental in designing industrial processes, where thermal management is essential.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When \(23.6 \mathrm{~g}\) of calcium chloride, \(\mathrm{CaCl}_{2}\), was dissolved in water in a constant-pressure calorimeter, the temperature rose from \(25.0^{\circ} \mathrm{C}\) to \(38.7^{\circ} \mathrm{C}\). If the heat capacity of the solution and the calorimeter is \(1258 \mathrm{~J} /{ }^{\circ} \mathrm{C}\), what is the enthalpy change when \(0.710 \mathrm{~mol}\) of calcium chloride dissolves in water? The solution process is $$ \mathrm{CaCl}_{2}(s) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) $$

Phosphoric acid, \(\mathrm{H}_{3} \mathrm{PO}_{4},\) can be prepared by the reaction of phosphorus(V) oxide, \(\mathrm{P}_{4} \mathrm{O}_{10},\) with water. \(\frac{1}{4} P_{4} \mathrm{O}_{10}(s)+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q) ; \Delta H=-96.2 \mathrm{~kJ}\) What is \(\Delta H\) for the reaction involving \(1 \mathrm{~mol}\) of \(\mathrm{P}_{4} \mathrm{O}_{10} ?\) $$ \mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(a q) $$

Calculate the grams of oxygen gas required to produce \(7.60 \mathrm{~kJ}\) of heat when hydrogen gas burns at constant pressure to give liquid water, given the following: $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) ; \Delta H=-484 \mathrm{~kJ} $$ Liquid water has a heat of vaporization of \(44.0 \mathrm{~kJ}\) per mole at \(25^{\circ} \mathrm{C}\)

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), is a colorless liquid whose solutions are used as a bleach and an antiseptic. \(\mathrm{H}_{2} \mathrm{O}_{2}\) can be prepared in a process whose overall change is $$ \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(l) $$ Calculate the enthalpy change using the following data: $$ \begin{array}{l} \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g) ; \Delta H=-98.0 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) ; \Delta H=-571.6 \mathrm{~kJ} \end{array} $$

You wish to heat water to make coffee. How much heat (in joules) must be used to raise the temperature of \(0.180 \mathrm{~kg}\) of tap water (enough for one cup of coffee) from \(30^{\circ} \mathrm{C}\) to \(96^{\circ} \mathrm{C}\) (near the ideal brewing temperature)? Assume the specific heat is that of pure water, \(4.18 \mathrm{~J} /\) \(\left(g \cdot{ }^{\circ} \mathrm{C}\right)\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.