Solution Found!
Diamond and graphite are two crystalline forms of carbon.
Chapter 5, Problem 48P(choose chapter or problem)
Diamond and graphite are two crystalline forms of carbon. At 1 atm and \(25^{\circ} \mathrm{C}\), diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Determine \(\Delta \mathrm{H}_{\text {rxn }}\)
\(\mathrm{C} \text { (diamond) } \rightarrow \mathrm{C} \text { (graphite) }\)
with equations from the following list:
\(\mathrm{C} \text { (diamond) }+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) \quad \Delta \mathrm{H}=-395.4 \mathrm{~kJ}\)
\(2 \mathrm{CO}_{2}(\mathrm{~g}) 2 \mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{O}_{2}(\mathrm{~g}) \quad \Delta \mathrm{H}=566.0 \mathrm{~kJ}\)
\(\mathrm{C}(\text { graphite })+\mathrm{O} 2(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) \quad \Delta \mathrm{H}=-393.5 \mathrm{~kJ}\)
\(2 \mathrm{CO}(\mathrm{g})+\mathrm{C}(\text { graphite }) \rightarrow \mathrm{CO} 2(\mathrm{~g}) \quad \Delta \mathrm{H}=-172.5 \mathrm{~kJ}\)
Equation Transcription:
25°C
Hrxn
C(diamond) C(graphite)
C(diamond) + O2(g) CO2(g)
H = -395.4 kJ
2CO2(g) 2CO(g) O2(g)
H = 566.0 kJ
C(graphite) + O2(g) CO2(g)
H = -393.5 kJ
2CO(g) + C(graphite) CO2(g)
H = -172.5 kJ
Text Transcription:
25°C
H_rxn
C(diamond) rightarrow C(graphite)
C(diamond) + O2(g) rightarrow CO2(g) Delta H = -395.4 kJ
2CO2(g) 2CO(g) rightarrow O2(g) Delta H = 566.0 kJ
C(graphite) + O2(g) rightarrow CO2(g) Delta H = -393.5 kJ
2CO(g) + C(graphite) rightarrow CO2(g) Delta H = -172.5 kJ
Questions & Answers
QUESTION:
Diamond and graphite are two crystalline forms of carbon. At 1 atm and \(25^{\circ} \mathrm{C}\), diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Determine \(\Delta \mathrm{H}_{\text {rxn }}\)
\(\mathrm{C} \text { (diamond) } \rightarrow \mathrm{C} \text { (graphite) }\)
with equations from the following list:
\(\mathrm{C} \text { (diamond) }+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) \quad \Delta \mathrm{H}=-395.4 \mathrm{~kJ}\)
\(2 \mathrm{CO}_{2}(\mathrm{~g}) 2 \mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{O}_{2}(\mathrm{~g}) \quad \Delta \mathrm{H}=566.0 \mathrm{~kJ}\)
\(\mathrm{C}(\text { graphite })+\mathrm{O} 2(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) \quad \Delta \mathrm{H}=-393.5 \mathrm{~kJ}\)
\(2 \mathrm{CO}(\mathrm{g})+\mathrm{C}(\text { graphite }) \rightarrow \mathrm{CO} 2(\mathrm{~g}) \quad \Delta \mathrm{H}=-172.5 \mathrm{~kJ}\)
Equation Transcription:
25°C
Hrxn
C(diamond) C(graphite)
C(diamond) + O2(g) CO2(g)
H = -395.4 kJ
2CO2(g) 2CO(g) O2(g)
H = 566.0 kJ
C(graphite) + O2(g) CO2(g)
H = -393.5 kJ
2CO(g) + C(graphite) CO2(g)
H = -172.5 kJ
Text Transcription:
25°C
H_rxn
C(diamond) rightarrow C(graphite)
C(diamond) + O2(g) rightarrow CO2(g) Delta H = -395.4 kJ
2CO2(g) 2CO(g) rightarrow O2(g) Delta H = 566.0 kJ
C(graphite) + O2(g) rightarrow CO2(g) Delta H = -393.5 kJ
2CO(g) + C(graphite) rightarrow CO2(g) Delta H = -172.5 kJ
ANSWER:
Solution 48P
Here we have to calculate Δ Hrxn for
C(diamond) → C(graphite)
Step 1
Given:
The following equations are given,
(1) C (diamond) + O2 (g) → CO2 (g) = -395.4 kJ
(2)2CO2(g) →2 CO (g) + O2(g) = 566.0 kJ
(3) C (graphite) + O2 (g) → CO2 (g) = -393.5 kJ
(4)2CO (g) → C (graphite) + CO2 (g) = -172.5 kJ
Here we will have to arrange all the equations in such a way that the final result must give C(diamond) → C(graphite). In order to do that we have to cancel the O2, CO and CO2.