281.1kJmol⁻¹;
−281.1kJmol⁻¹;
−562.2kJmol⁻¹;
562.2kJmol⁻¹;
Correct option is B. −281.1kJmol⁻¹
Given:-
C+O2→CO2 ΔH=−393kJ/mol
2C+2O2→2CO2 ΔH=−786kJ/mol
H2+1/2O2→H2O ΔH=−287.3kJ/mol
3H2+3/2O2→3H2O ΔH=−861.9kJ/mol
2CO2+3H2O→C2H5OH+3O2 ΔH=1366.8kJ/mol
Adding equation(1),(2) (3), we get-
2C+3H2+1/2O2→C2H5OH ΔH=−281.1kJ/mol
Hence, the standard enthalpy of formation of C2H5OH is −281.1kJ/mol.