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
General Chemistry - Standalone book (MindTap Course List)
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Chapter18: Thermodynamics And Equilibrium
Section: Chapter Questions
Problem 18.97QP: When 1.000 g of gaseous butane, C4H10, is burned at 25C and 1.00 atm pressure, H2O(l) and CO2(g) are...
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F52d4d029-378f-48d0-8332-5b17fae15160%2F4f09b0e8-b57b-4d4c-94c5-f04e9c19a035%2Fiamlyl_processed.jpeg&w=3840&q=75)
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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