Find ∆H as a function of T for the reaction: CO2 (g) + C (s, graphite) = CO Given that ∆H 293= 41,400 cal Cp CO2= 5.166+15.77×10^-3 T - 95.78×10^-3 T² + 2.260×10^-9 T³ Cp CO= 6.6 + 2×10^-3 T Cp C (s, graphite)= 2.673 + 2.617×10^-3 T - 1.169×10^-5 T^-2
Find ∆H as a function of T for the reaction: CO2 (g) + C (s, graphite) = CO Given that ∆H 293= 41,400 cal Cp CO2= 5.166+15.77×10^-3 T - 95.78×10^-3 T² + 2.260×10^-9 T³ Cp CO= 6.6 + 2×10^-3 T Cp C (s, graphite)= 2.673 + 2.617×10^-3 T - 1.169×10^-5 T^-2
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter16: Thermodynamics: Directionality Of Chemical Reactions
Section: Chapter Questions
Problem 8QRT
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Find ∆H as a function of T for the reaction: CO2 (g) + C (s, graphite) = CO
Given that ∆H 293= 41,400 cal
Cp CO2= 5.166+15.77×10^-3 T - 95.78×10^-3 T² + 2.260×10^-9 T³
Cp CO= 6.6 + 2×10^-3 T
Cp C (s, graphite)= 2.673 + 2.617×10^-3 T - 1.169×10^-5 T^-2
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