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Given enthalpy of formation of \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{CaO}(\mathrm{s})\) are \(-94.0 \mathrm{~kJ}\) and \(-152 \mathrm{~kJ}\), respectively, and the enthalpy of the reaction: \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) is \(42 \mathrm{~kJ}\). The enthalpy of formation of \(\mathrm{CaCO}_{3}(\mathrm{~s})\) is (a) \(-42 \mathrm{~kJ}\) (b) \(-202 \mathrm{~kJ}\) (c) \(+202 \mathrm{~kJ}\) (d) \(-288 \mathrm{~kJ}\)

Short Answer

Expert verified
-202 \mathrm{~kJ}

Step by step solution

01

Understand the Hess's Law

According to Hess's Law, the change in enthalpy for a chemical reaction is the same, no matter how many steps the reaction is carried out in. This means the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps.
02

Apply Hess's Law to calculate the enthalpy of formation of \(\text{CaCO}_3(\text{s})\)

The enthalpy of the given reaction \(\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})\) is the sum of the enthalpies of formation of the products minus the enthalpy of formation of the reactant (\text{CaCO}_3(\text{s})). Using the reaction enthalpy and the given enthalpies of formation for \text{CO}_2(\text{g}) and \text{CaO}(\text{s}), we can find the enthalpy of formation for \text{CaCO}_3(\text{s}).
03

Solve for the enthalpy of formation of \(\text{CaCO}_3(\text{s})\)

The enthalpy of formation of \text{CaCO}_3(\text{s}) can be found using the equation: \[\text{Enthalpy of formation of \text{CaCO}_3(\text{s})} = \text{Enthalpy of reaction} + \text{Enthalpy of formation of \text{CaO}(\text{s})} + \text{Enthalpy of formation of \text{CO}_2(\text{g})}\]. Substitute the known values: \[\text{Enthalpy of formation of \text{CaCO}_3(\text{s})} = 42 \text{kJ} - 152 \text{kJ} - (-94 \text{kJ}) \] Solve for the enthalpy of formation of \text{CaCO}_3(\text{s}).

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Key Concepts

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

Understanding Hess's Law
At its core, Hess's Law is a principle in chemical thermodynamics offering a straightforward way to calculate the enthalpy change of a reaction. It states that the total enthalpy change during the complete course of a chemical reaction is the same, whether the reaction is made in one step or several steps.

Hess's Law is incredibly practical because it allows chemists to use tabulated enthalpy values of known reactions (called enthalpies of formation) to predict the enthalpy change of reactions where direct measurement may be difficult. Essentially, if you can write an unknown reaction as the sum of a series of steps for which you know the enthalpy changes, you can accurately calculate the enthalpy change for the reaction of interest.

To apply Hess's Law to solve a problem, you should:
  • Define the final reaction of interest and, if necessary, break it down into known steps.
  • Ensure each step’s equation is correctly written with states of matter and coefficients.
  • Use the standard enthalpy of formation data for the substances involved.
  • Add up the enthalpy changes of each step to find the total reaction enthalpy.
The Basics of Chemical Thermodynamics
Preparing to dive into a topic like chemical thermodynamics requires a grasp of its fundamental concepts. Chemical thermodynamics deals with the study of energy changes, particularly the concept of energy conservation and the interplay of energy in chemical reactions called enthalpy.

In the context of enthalpy change, thermodynamics seeks to explain how heat is absorbed or released by a system during chemical processes. This energy change, expressed in units of joules or kilojoules, is an essential aspect of reactants and products in chemical reactions, determining whether a process is endothermic (absorbing heat) or exothermic (releasing heat).

A central rule in chemical thermodynamics is that every chemical reaction strives to reach a state of lowest energy and highest entropy, which translates to the system's spontaneity and stability. Understanding this concept aids in predicting the feasibility and direction of chemical reactions.
Enthalpy Change Explained
Enthalpy change itself is a measure of heat change at constant pressure and is signified by the symbol ΔH. As a critical quantity in chemical thermodynamics, it details whether energy is absorbed or emitted during a reaction.

Positive ΔH values signal endothermic reactions, where energy is absorbed from the surroundings. On the contrary, exothermic reactions have negative ΔH values, indicating energy release. The standard enthalpy of formation, or ΔH°f, is referenced when one mole of a substance is formed from its constituent elements in their standard states.

In connection with our exercise, using enthalpy changes for known substances such as CO2(g) and CaO(s), and applying Hess's Law, one can deduce unknown values such as the standard enthalpy of formation for CaCO3(s) by accounting for the energy change during its decomposition. This knowledge is paramount when predicting the behavior of a substance during a reaction or in the environment.

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Most popular questions from this chapter

A geyser, operating on LPG (liquefied petroleum gas) heats water flowing at the rate of \(3.0\) litres per minutes, from \(27^{\circ} \mathrm{C}\) to \(77^{\circ} \mathrm{C}\). If the heat of combustion of LPG is \(40,000 \mathrm{~J} / \mathrm{g}\), how much fuel, in \(\mathrm{g}\), is consumed per minute? (Specific heat capacity of water is \(4200 \mathrm{~J} / \mathrm{kg}-\mathrm{K}\) ) (a) \(15.25\) (b) \(15.50\) (c) \(15.75\) (d) \(16.00\)

The enthalpies of formation of \(\mathrm{FeO}(\mathrm{s})\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})\) are \(-65.0\) and \(-197.0 \mathrm{kcal} /\) mol, respectively. A mixture of the two oxides contains \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the mole ratio \(2: 1 .\) If by oxidation it is changed in to a \(1: 2\) mole ratio mixture, how much of thermal energy will be released per mole of the initial mixture? (a) \(13.4 \mathrm{kcal}\) (b) \(67 \mathrm{kcal}\) (c) \(47.2 \mathrm{kcal}\) (d) 81 kcal

The value of \(\Delta H_{\text {sol }}\) of anhydrous \(\begin{array}{lllll}\text { copper (II) sulphate } & \text { is } & -66.11 & \mathrm{~kJ}\end{array}\) Dissolution of 1 mole of blue vitriol, [Copper (II) sulphate pentahydrate] is followed by absorption of \(11.5 \mathrm{~kJ}\) of heat. The enthalpy of dehydration of blue vitriol is (a) \(-77.61 \mathrm{~kJ}\) (b) \(+77.61 \mathrm{~kJ}\) (c) \(-54.61 \mathrm{~kJ}\) (d) \(+54.61 \mathrm{~kJ}\)

Enthalpies of solution of \(\mathrm{BaCl}_{2}(\mathrm{~s})\) and \(\mathrm{BaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) are \(-20.6 \mathrm{~kJ} / \mathrm{mol}\) and \(8.8 \mathrm{~kJ} / \mathrm{mol}\), respectively. \(\Delta H\) hydration of \(\mathrm{BaCl}_{2}(\mathrm{~s})\) to \(\mathrm{BaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) is (a) \(-29.4 \mathrm{~kJ}\) (b) \(-11.8 \mathrm{~kJ}\) (c) \(29.6 \mathrm{~kJ}\) (d) \(11.8 \mathrm{~kJ}\)

The standard enthalpies of formation of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l}), \mathrm{Li}^{+}(\mathrm{aq})\) and \(\mathrm{OH}^{-}(\mathrm{aq})\) are \(-285.8\), \(-278.5\) and \(-228.9 \mathrm{~kJ} / \mathrm{mol}\), respectively. The standard enthalpy change for the reaction is \(2 \mathrm{Li}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{Li}^{+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})\) \(+\mathrm{H}_{2}(\mathrm{~g})\) (a) \(+443.2 \mathrm{~kJ}\) (b) \(-443.2 \mathrm{~kJ}\) (c) \(-221.6 \mathrm{~kJ}\) (d) \(+221.6 \mathrm{~kJ}\)

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