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Calculate the standard enthalpy of formation of phenol, C 6H50H, at 298.15 K given that, at this temperature, the standard enthalpy of formation o phenol is -165.0 kJ mol, of liquid water, H2O is -285.8 kJ mol and gaseous carbon dioxide, CO2, is -393.51 kJ mol1 2C6H5OH + 1502 12CO2 + 6H2O O a. -2202.9 kJ mol-1 O b.-844.3 kJ mol1 O c. -3053.5 kJ mol-1 O d. 514.3 kJ mol-1

Calculate the standard enthalpy of formation of phenol, C 6H50H, at 298.15 K given that, at this temperature, the standard enthalpy of formation o phenol is -165.0 kJ mol, of liquid water, H2O is -285.8 kJ mol and gaseous carbon dioxide, CO2, is -393.51 kJ mol1 2C6H5OH + 1502 12CO2 + 6H2O O a. -2202.9 kJ mol-1 O b.-844.3 kJ mol1 O c. -3053.5 kJ mol-1 O d. 514.3 kJ mol-1

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Calculate the standard enthalpy of formation of phenol, C 6H5OH, at 298.15 K given that, at this temperature, the standard enthalpy of formation o phenol is -165.0 kJ mol, of liquid water, H2O is -285.8 kJ mol and gaseous carbon dioxide, CO2, is -393.51 kJ mol-1 2C6H5OH + 1502 → 12CO2 + 6H20 O a. -2202.9 kJ mol1 O b.-844.3 kJ mol1 O c. -3053.5 kJ mol-1 O d. 514.3 kJ mol-1
Transcribed Image Text:Calculate the standard enthalpy of formation of phenol, C 6H5OH, at 298.15 K given that, at this temperature, the standard enthalpy of formation o phenol is -165.0 kJ mol, of liquid water, H2O is -285.8 kJ mol and gaseous carbon dioxide, CO2, is -393.51 kJ mol-1 2C6H5OH + 1502 → 12CO2 + 6H20 O a. -2202.9 kJ mol1 O b.-844.3 kJ mol1 O c. -3053.5 kJ mol-1 O d. 514.3 kJ mol-1
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