Chapter 6: Problem 53
\(C_{\text {(dianemd) }}+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) ; \Delta H=395 \mathrm{~kJ}\) \(C_{\text {(Braphite) }}+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) ; \Delta / I=-393.5 \mathrm{~kJ}\) The \(\Delta / /\) when diamond is formed from graphite (1) \(-1.5 \mathrm{~kJ}\) \((2)+1.5 \mathrm{~kJ}\) (3) \(+3.0 \mathrm{k} \mathrm{J}\) (4) \(-3.0 \mathrm{~kJ}\)
Short Answer
Expert verified
The enthalpy change when diamond is formed from graphite is +1.5 kJ.
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
enthalpy calculation
Enthalpy is a measure of total energy in a thermodynamic system. It's essential in understanding energy changes during chemical reactions.
For enthalpy calculation, we look at the change in enthalpy (\( \triangle H \)) between reactants and products. If the reaction absorbs heat, \(\triangle H\) is positive, indicating an endothermic reaction. Conversely, if it releases heat, \(\triangle H\) is negative, signifying an exothermic reaction.
In the example provided, we calculate the enthalpy difference for the formation of diamond from graphite. We first note the enthalpy values for the given reactions. Then, we find the enthalpy change by subtracting the enthalpy of the graphite reaction from the diamond reaction: \(\triangle H_{\text{target}} = \triangle H_{\text{(diamond)}} - \triangle H_{\text{(graphite)}}\). This yields \(\triangle H_{\text{target}} = 395 \text{ kJ} - (-393.5 \text{ kJ}) = 1.5 \text{ kJ}\).
For enthalpy calculation, we look at the change in enthalpy (\( \triangle H \)) between reactants and products. If the reaction absorbs heat, \(\triangle H\) is positive, indicating an endothermic reaction. Conversely, if it releases heat, \(\triangle H\) is negative, signifying an exothermic reaction.
In the example provided, we calculate the enthalpy difference for the formation of diamond from graphite. We first note the enthalpy values for the given reactions. Then, we find the enthalpy change by subtracting the enthalpy of the graphite reaction from the diamond reaction: \(\triangle H_{\text{target}} = \triangle H_{\text{(diamond)}} - \triangle H_{\text{(graphite)}}\). This yields \(\triangle H_{\text{target}} = 395 \text{ kJ} - (-393.5 \text{ kJ}) = 1.5 \text{ kJ}\).
chemical thermodynamics
Chemical thermodynamics focuses on the energy changes (enthalpy, entropy, Gibbs free energy) occurring in chemical processes.
It involves the study of laws governing these energy transformations, particularly the First Law of Thermodynamics which states that energy cannot be created or destroyed. Thus, the total energy of the universe is constant. When applied to our enthalpy change problem, it helps to understand how energy is conserved and transferred in reactions.
In the given reactions, the overall enthalpy change reflects how much energy is absorbed or released during the transformation of graphite to diamond. As we calculated, this transition absorbs 1.5 kJ of energy, emphasizing that diamond forms under specific thermodynamic conditions.
It involves the study of laws governing these energy transformations, particularly the First Law of Thermodynamics which states that energy cannot be created or destroyed. Thus, the total energy of the universe is constant. When applied to our enthalpy change problem, it helps to understand how energy is conserved and transferred in reactions.
In the given reactions, the overall enthalpy change reflects how much energy is absorbed or released during the transformation of graphite to diamond. As we calculated, this transition absorbs 1.5 kJ of energy, emphasizing that diamond forms under specific thermodynamic conditions.
formation reactions
Formation reactions involve creating one mole of a compound from its elements in their standard states.
For example, forming \(\text{CO}_2\) from graphite involves combining carbon in its standard form (graphite) with oxygen gas: \(\text{C}_{\text{(graphite)}} + \text{O}_2 (\text{g}) \rightarrow \text{CO}_2 (\text{g}) \).
The enthalpy change of this reaction tells us the amount of energy released or absorbed when forming \(\text{CO}_2\). Similarly, for diamond: \(\text{C}_{\text{(diamond)}} + \text{O}_2 (\text{g}) \rightarrow \text{CO}_2 (\text{g}) \).
By comparing these reactions, we discern the energy difference between diamond and graphite. Calculating \(\triangle H_{\text{target}}\) signifies this energy gap, already found to be 1.5 kJ.
For example, forming \(\text{CO}_2\) from graphite involves combining carbon in its standard form (graphite) with oxygen gas: \(\text{C}_{\text{(graphite)}} + \text{O}_2 (\text{g}) \rightarrow \text{CO}_2 (\text{g}) \).
The enthalpy change of this reaction tells us the amount of energy released or absorbed when forming \(\text{CO}_2\). Similarly, for diamond: \(\text{C}_{\text{(diamond)}} + \text{O}_2 (\text{g}) \rightarrow \text{CO}_2 (\text{g}) \).
By comparing these reactions, we discern the energy difference between diamond and graphite. Calculating \(\triangle H_{\text{target}}\) signifies this energy gap, already found to be 1.5 kJ.
