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Calculate the standard enthalpy of formation for diamond, given that $$ \begin{aligned} \text { C(graphite) }+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta H^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{C}(\text { diamond })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \\ \Delta H^{\circ} &=-395.4 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$

Short Answer

Expert verified
The standard enthalpy of formation for diamond is \(-1.9 \, \text{kJ/mol}\).

Step by step solution

01

Understand the Problem

We are tasked with calculating the standard enthalpy of formation for diamond from graphite and oxygen. We are given the enthalpies for the combustion of both graphite and diamond to form carbon dioxide.
02

Write the Known Reactions

The given reactions are: 1. \( \text{C(graphite) + O}_2(g) \rightarrow \text{CO}_2(g) \) with \( \Delta H^{\circ} = -393.5 \, \text{kJ/mol} \). 2. \( \text{C(diamond) + O}_2(g) \rightarrow \text{CO}_2(g) \) with \( \Delta H^{\circ} = -395.4 \, \text{kJ/mol} \).
03

Determine the Target Reaction

We need the enthalpy change for the following reaction: \( \text{C(graphite)} \rightarrow \text{C(diamond)} \).
04

Apply Hess's Law

According to Hess's Law, the change in enthalpy for a reaction is the same regardless of the pathway. We can use this principle to determine the enthalpy change for converting graphite to diamond. Subtract the enthalpy of combustion of graphite from that of diamond: \[ \Delta H_{\text{formation, diamond}} = \Delta H_{\text{C(diamond)}} - \Delta H_{\text{C(graphite)}} \]
05

Perform the Calculation

Substitute the given values into the equation derived from Hess's Law:\[ \Delta H_{\text{formation, diamond}} = (-395.4) - (-393.5) = -395.4 + 393.5 = -1.9 \, \text{kJ/mol} \]
06

Conclude the Result

The calculated standard enthalpy of formation for diamond from graphite is \( -1.9 \, \text{kJ/mol} \).

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Key Concepts

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

Hess's Law
Hess's Law is a crucial principle in chemistry used to determine the enthalpy change for a chemical reaction that can be expressed as the sum of other reactions. This law states that the total enthalpy change for a reaction is the same, no matter the series of steps you take.
Hess's Law is based on the first law of thermodynamics, which implies that enthalpy is a state function. Because enthalpy is a state function, the path taken to achieve the final state is irrelevant; only the initial and final states matter.
For example, if you have reactions A to B and C to D, Hess’s Law allows you to determine the enthalpy change for a reaction involving A and D by a linear combination of these known reactions.
In our exercise, Hess's Law helps us find the unknown enthalpy change for the conversion of graphite to diamond using the given combustion reactions.
Enthalpy Change
Enthalpy change, represented as \( \Delta H \), is the heat absorbed or released during a chemical reaction carried out at constant pressure. It provides crucial insight into the energy dynamics of chemical processes.
An enthalpy change can be either exothermic or endothermic.
  • Exothermic reactions release heat, resulting in a negative enthalpy change (\( \Delta H < 0 \)).
  • Endothermic reactions absorb heat, leading to a positive enthalpy change (\( \Delta H > 0 \)).
The magnitude of this change is significant in understanding the energy required or released in chemical processes, essential to predicting reaction behaviors.
In the given exercise, we calculate the enthalpy change for converting graphite to diamond by subtracting the enthalpy of graphite's combustion from that of diamond.
Combustion Reactions
Combustion reactions involve the reaction of a substance with oxygen to produce heat and often light. In these reactions, an element or compound swiftly combines with oxygen to form oxides.
Combustion can be a complete reaction, forming water and carbon dioxide, typically when sufficient oxygen is available.
  • Graphite's combustion: \( \text{C(graphite) + O}_2(g) \rightarrow \text{CO}_2(g) \)
  • Diamond's combustion: \( \text{C(diamond) + O}_2(g) \rightarrow \text{CO}_2(g) \)
Both reactions release large amounts of heat, as indicated by their negative enthalpy values.
For example, the combustion enthalpy for graphite is \( -393.5 \, \text{kJ/mol} \), while for diamond, it is \( -395.4 \, \text{kJ/mol} \).
These reactions are used in the exercise to deduce the enthalpy change from graphite to diamond through Hess's Law.
Diamond Formation
Diamond formation involves converting carbon from one structure, graphite, into another, diamond. Despite both being allotropes of carbon, they exhibit different lattice structures.
Diamonds have a three-dimensional tetrahedral structure, making them extremely strong and resistant. This transformation is intriguing because it involves a change in carbon's atomic arrangement without altering its chemical identity.
The conditions for naturally forming diamonds from graphite are extreme, typically requiring high pressure and temperature, conditions present deep within Earth's mantle.
In our exercise, we explore the energetic aspect of converting graphite to diamond by calculating the standard enthalpy of formation for diamond. This is effectively done by using known combustion reactions and applying Hess's Law.
Graphite Conversion
Graphite conversion primarily focuses on the transformation of graphite into diamond, focusing on its thermodynamic aspects. Graphite, another carbon allotrope, has a layered, planar structure, which makes it soft and slippery.
The conversion to diamond involves an energy change due to the rearrangement of carbon atoms.
In the exercise, graphite conversion is analyzed to find the enthalpy change using Hess's Law. By determining the energy difference between the combustion of graphite and diamond, this enthalpy change represents the energy associated with converting graphite into diamond.
This transformation highlights graphite's dynamic chemistry and its potential to alter its structure under certain conditions, showcasing the fascinating adaptability of carbon structures.

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Most popular questions from this chapter

The first step in the industrial recovery of zinc from the zinc sulfide ore is roasting; that is, the conversion of \(\mathrm{ZnS}\) to \(\mathrm{ZnO}\) by heating:$$\begin{aligned}2 \mathrm{ZnS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{ZnO}(s)+2 \mathrm{SO}_{2}(g) & \Delta H=-879 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$$ Calculate the heat evolved (in kJ) per gram of \(\mathrm{ZnS}\) roasted.

A 46-kg person drinks \(500 \mathrm{~g}\) of milk, which has a "caloric" value of approximately \(3.0 \mathrm{~kJ} / \mathrm{g}\). If only 17 percent of the energy in milk is converted to mechanical work, how high (in meters) can the person climb based on this energy intake?

Lime is a term that includes calcium oxide \((\mathrm{CaO},\) also called quicklime) and calcium hydroxide \(\left[\mathrm{Ca}(\mathrm{OH})_{2}\right.\) also called slaked lime]. It is used in the steel industry to remove acidic impurities, in air-pollution control to remove acidic oxides such as \(\mathrm{SO}_{2}\), and in water treatment. Quicklime is made industrially by heating limestone \(\left(\mathrm{CaCO}_{3}\right)\) above \(2000^{\circ} \mathrm{C}:\) \(\begin{aligned} \mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) & \Delta H^{\circ}=177.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned}\) Slaked lime is produced by treating quicklime with water: \(\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)_{\Delta H^{\circ}}=-65.2 \mathrm{~kJ} / \mathrm{mol}\) The exothermic reaction of quicklime with water and the rather small specific heats of both quicklime \(\left[0.946 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\right]\) and slaked lime \(\left[1.20 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\right]\) make it hazardous to store and transport lime in vessels made of wood. Wooden sailing ships carrying lime would occasionally catch fire when water leaked into the hold. (a) If a 500.0 -g sample of water reacts with an equimolar amount of \(\mathrm{CaO}\) (both at an initial temperature of \(\left.25^{\circ} \mathrm{C}\right)\), what is the final temperature of the product, \(\mathrm{Ca}(\mathrm{OH})_{2} ?\) Assume that the product absorbs all the heat released in the reaction. (b) Given that the standard enthalpies of formation of \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O}\) are -635.6 and \(-285.8 \mathrm{~kJ} / \mathrm{mol}\), respectively, calculate the standard enthalpy of formation of \(\mathrm{Ca}(\mathrm{OH})_{2}\).

In writing thermochemical equations, why is it important to indicate the physical state (i.e., gaseous, liquid, solid, or aqueous) of each substance?

A driver's manual states that the stopping distance quadruples as the speed doubles; that is, if it takes \(30 \mathrm{ft}\) to stop a car moving at \(25 \mathrm{mph}\), then it would take \(120 \mathrm{ft}\) to stop a car moving at \(50 \mathrm{mph}\). Justify this statement by using mechanics and the first law of thermodynamics. (Assume that when a car is stopped, its kinetic energy \(\left(\frac{1}{2} m u^{2}\right)\) is totally converted to heat.)

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